(A) Angular velocity upwards, angular acceleration downwards.

(C) Both the statements are correct.

(B) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.

(B) 2.5

(C) For both, (P) and (Q)

(D) 2:1:3

To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road.

On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.

• When a two-wheeler a turn along an unbanked road, the force of friction provides the centripetal force.
• The two wheeler leans inward to counteract a torque that tends to topple it outward.
• Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions.
• Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.

The frequency of conical pendulum of string length l and semi vertical angle q is

n = \(\frac{1}{2π}\sqrt[g}{L.cosθ}\)

where g is the acceleration due to gravity at the place

from above expression we can say that

(i) n ∝ \(\sqrt{g}\)

(ii) n  ∝ \(\frac{1}{\sqrt{L}}\)

(iii) n ∝   \(\frac{1}{cosθ}\) ….If θ increases cosθ decreases and n increases

(iv) The frequency is independent of the mass of the bob.

Radius of gyration K =\(\sqrt{\frac{I}{M}}\)

The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

The moment of inertia of a hollow right circular cone about its central symmetry axis is

I = (1/2)MR²,

The same as that of a disc about its transverse symmetry axis. This is because the distribution of mass of the hollow cone about its central symmetry axis is the same as that of a disc.

(i) When a bicyclist takes a turn along an unbanked road, the force of friction \(\vec{f_s}\)   provides the centripetal force; the normal reaction of the road \(\vec{N}\)  is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, f_s·h, due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by  and the weight g,

mg·a = f_s·h₁.

(ii) Since the force of friction provides the centripetal force,

\(f_s=\frac{mv^2}{r}\)

If the cyclist leans from the vertical by an angle q, the angle between \(\vec{N}\) and \(\vec{F}\) as shown in fig (b).

tanθ=\(\frac{f_s}{N}\ = \frac{mv^2/r}{mg}=\frac{v^2}{gr}\)

Hence, the cyclist must lean by an angle

θ =tan^-1 \(\frac{v^2}{gr}\)

(iii) When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels, So, when the inner wheels just pet lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.

Consider a particle of mass m attached to a string and revolved in vertical circle of radius r, At every instant of motion  there are only two forces acting on the particle (a) its weight mg, vertically downwards, which is constant and

(b) the force due to the tension along the string, directed along the string and towards the centre.

Its magnitude changes periodically with time and location.

The particle may not complete the circle if the string slackens before the particle reaches at the top. This requires that the particle must have some minimum speed.

At the top position (Point A) : Let v_a be the speed of the particle and T_A the tension in the string. Here both, weight mg and force due to tension T_A are downwards, i.e. towards the centre. In this case, net force on the particle towards the center O is T_A + mg is their resultant as the centripetal force.

∴ T_A +mg = \(\frac{m{v_A}^2}{r}\)   ……..(1)

For minimum possible speed at this point (or if the motion is to be realized with minimum possible energy),

T_A = 0  

∴ 0 + mg =  \(\frac{m{v_A}^2}{r}\)

∴   \(\frac{m{v_A}^2}{r}\) = mg

That is the particle’s weight alone is the necessary centripetal force at the point A.

∴ v_A² = rg

∴ v_A =  \(\sqrt{rg}\) ……..(2)

At the bottom position (Point B): Let v_B The speed at bottom. Taking reference level for zero potential energy to the bottom of the circle, the particle has only kinetic energy \(\frac{1}{2}m{v_B}^2\)  at the lowest point.

Total energy at the bottom = KE + PE

= \(\frac{1}{2}m{v_B}^2\)+ 0 = \(\frac{1}{2}m{v_B}^2\)   …… (3)

As the particle goes from the bottom to the top of the circle, it rises through a height h=2r Therefore potential energy at top is

mgh=mg(2r) and from eq. 2 its minimum KE there is

\(\frac{1}{2}m{v_A}^2\)= \(\frac{1}{2}mgr\) …… (4)

Minimum Total energy at the top = KE+PE

= \(\frac{1}{2}mgr + 2mgr\)= \(\frac{5}{2}mgr\)…… (5)

Assuming that the total energy of the particle is conserved,

Total energy at the bottom = Total energy at the top.

From eq. 4 & 5

\(\frac{1}{2}m{v_B}^2\) = \(\frac{5}{2}mgr\)

The minimum speed the particle must have at the bottom is

v_B =\(\sqrt{5gr}\)  …….. (6)

(iii) At the midway (Point C)

Let v_C be the speed at point C, so that the kinetic energy is \(\frac{1}{2}m{v_C}^2\)

At C the particle is at height r from the bottom of the circle.

Therefor PE at C is mgr

Total Energy at C

= \(\frac{1}{2}m{v_C}^2 + mgr\)……. (7)

From the law of conservation of energy total energy at C = total energy at B

∴ \(\frac{1}{2}m{v_C}^2 + mgr\) = \(\frac{5}{2}mgr\)

∴ v_C² = 5gr-2gr = 3gr

∴ v_C = \(\sqrt{3gr}\)

In a nonuniform vertical circular motion, e.g., those of a small body attached to a string or the loop maneuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity and zero speed at the top is not possible.

However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.

Consider a small body (or particle) of mass m tied to a string and revolved in a vertical circle of radius rat a place where the acceleration due to gravity is g. At every instant of its motion, the body is acted upon by two forces, namely, its weight mg and the tension T in the string.

Let v_B be the speed of the body and T_B be the tension in the string at the lowest point B. We take the reference level for zero potential energy to be the bottom of the circle. Then, the body has only kinetic energy  at the lowest point.

T_B = \(\frac{m{v_B}^2}{r}+mg\) .........(1)

Total energy at the bottom is KE+PE

= \(\frac{1}{2}m{v_B}^2+0\)

= \(\frac{1}{2}m{v_B}^2\) ..........(2)

Let v_A be the speed of the body and T_A be the tension in the string at the lowest point A. As the body goes from B to A it rises through a height h=2r

∴ T_A = \(\frac{m{v_A}^2}{r}-mg\) .........(3)

Total energy at A = KE+PE

= \(\frac{m{v_A}^2}{r}+mg(2r)\) .........(4)

Then from eq. (1) and (3)

T_B - T_A = \(\frac{m{v_B}^2}{r}+mg- \frac{m{v_A}^2}{r}-mg\)

∴ T_B - T_A = \(\frac{m}{r}({v_B}^2-{v_A}^2) +2mg\) .........(5)

Assuming that the total energy of the body is conserved,

Total energy at the Bottom=Total energy at Top

From eq. (2) and (4)

\(\frac{1}{2}m{v_B}^2\)= \(\frac{m{v_A}^2}{r}+mg(2r)\)

v_B²-v_A² = 4gr .........(6)

Substituting this in eq. (5)

T_B - T_A = \(\frac{m}{r}(4gr) +2mg\)

∴ T_B - T_A  = 4mg+2mg = 6mg

Therefore difference in the tensions in the string at the highest and the lowest point is 6 times the weight of the body.

Definition: The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.

The moment of inertia (MI) of a body about a given rotation axis depends upon

(i) the mass of the body and

(ii) the distribution of mass about the axis of rotation.

These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above.

Thus,

I= \(\sum \limits_{i} m_ir^2_i\)

∴ k= \(\sqrt\frac{I}{M}\)

Physical significance : The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.

The centre of mass (CM) coordinates locates a point where if the entire mass M of a system of particles or that of a rigid body can be thought to be concentrated such that the acceleration of this point mass obeys Newton’s second law of motion, viz.

F = Ma_CM, where F is the sum of all the external forces acting on the body or on the individual particles of the system of particles.

Similarly, radius of gyration locates a point from the axis of rotation where the entire mass M can be thought to be concentrated such that the angular acceleration of that point mass about the axis rotation obeys the relation, τ = Mα , where τ is the sum of all the external torques acting on the body or on the individual particles of the system particles.

Consider a uniform ring and a uniform disc, both of the same mass M and same radius R.

Let I_r and I_d be their respective moment of inertias.

If K_r and K_d are their respective radii of gyration, we can write,

I_r = MR² = MK_r²

∴K_r = R and

I_d = (1/2) MR² = MK_d²

∴ K_d = R/ \(\sqrt{2}\)

∴ K_d < K_r

It shows mathematically that K is decided by the distribution of mass.

In a ring the entire mass is distributed at the distance R,

While for a disc, its mass is distributed between 0 and R.

Among any objects of same mass and radius, ring has the largest radius of gyration and hence maximum M.I.

The theorem of parallel axis is applicable to any body of arbitrary shape. The moment of inertia (MI) of the body about an axis through the centre mass should be known, say, I_CM. Then, the theorem can be used to find the MI, I, of the body about an axis parallel to the above axis. If the distance between the two axes is h,

I= I_CM + Mh²

The theorem of perpendicular axes is applicable to a plane lamina only. The moment of inertia of a plane lamina about an axis—the z axis-perpendicular to its plane is equal to the sum of its moments of inertia I_x and I_y about two mutually perpendicular axes x and y in its plane and through the point of intersection of the perpendicular axis and the lamina.

I_z=I_x+I_y

Consider a rigid body rotating with a constant angular velocity w about an axis through the point O and perpendicular to the plane of the figure. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity ω

Suppose that the body consists of N particles of masses m₁, m₂, ..., m_n, situated at perpendicular distances r₁, r₂, ..., r_n, respectively from the axis of rotation.

The particle of mass m, revolves along a circle of radius r,, with a linear velocity of magnitude

v₁= r₁ω The magnitude of the linear momentum of the particle is

p₁ = m₁v₁ = m₁r₁ω

The angular momentum of the particle about the axis of rotation is by definition,

\(\vec{L_1}=\vec{r_1}\vec{p_1}\)

\ L₁ = r₁ p₁sinq

where θ is the smaller of the two angles between r₁ and p₁

In this case, θ = 90° ∴ sinθ = 1

∴ L₁ = r₁ p_{1 =} r₁m₁r₁ω = m₁r₁²ω

Similarly, L₂ = m₂r₂²ω, L₃ = m₃r₃²ω etc.

The angular momentum of the body about the given axis is

L= L₁ + L₂ +L_{3 …….} + L_N

= m₁r₁²ω + m₂r₂²ω + m₃r₃²ω …….. m_Nr_N²ω

= ω (m₁r₁² + m₂r₂² + m₃r₃² …….. m_Nr_N²)

= \((\sum \limits_{i=1}^{N}m_ir_i^2)ω\)

∴ L=I ω

where I = \((\sum \limits_{i=1}^{N}m_ir_i^2)\)

= moment of inertia of the body about the given axis.

In vector form, \(\vec{L}=I\vec{ω}\)

Thus, angular momentum = moment of inertia x angular velocity.

Note: Angular momentum is a vector quantity. It has the same direction as \(\vec{ω}\)

A torque acting on a body produces angular acceleration.

Figure shows a rigid object rotating with a constant angular acceleration a about an axis perpendicular to the plane of paper.

For theoretical simplification let us consider the object to be consisting of N number of particles of masses m₁, m₂ , ..... m_N

at respective perpendicular distances r₁, r₂,.....r_N from the axis of rotation.

As the object rotates, all these particles perform circular motion with same angular acceleration a, but with different linear (tangential) accelerations

a₁=r₁α, a₂=r₂α, a₃=r₃α, …….., a_N=r_Nα, etc.

Force experienced by the first particle is

f₁=m₁a₁= m₁r₁α,

As these forces are tangential, their respective perpendicular distances from the axis are r₁, r₂,.....r_N

Thus, the torque experienced by the first particle is of magnitude τ₁ =f₁r₁ = m₁r₁²α

Similarly,   τ₂ = m₂r₂²α, τ₃ =f₃r₃ = m₃r₃²α …….., τ_N =f_Nr₃ = m_Nr_N²α

If the rotation is restricted to a single plane, directions of all these torques are the same, and along the axis. Magnitude of the resultant torque is then given by

τ = τ₁ + τ₂ +τ₃ …………,  +τ_N

= α (m₁r₁² + m₂r₂² + m₃r₃² …….. m_Nr_N²) = Iα

where, I = m₁r₁² + m₂r₂² + m₃r₃² …….. m_Nr_N² is the moment of inertia of the object about the given axis of rotation.

This gives the required relation τ = Iα

Angular acceleration \(vec{α}\) has the same direction as torque \(\vec{t}\) and both them are axial vectors.

Law (or principle) of conservation of angular momentum : The angular momentum of a body is conserved if the resultant external torque on the body is zero.

Explanation : This law (or principle) is used by a figure skater or a ballerina to increase their speed of rotation for a spin by reducing the body’s moment of inertia. A diver too uses it during a somersault for the same reason.

(1) Ballet dancers: During ice ballet, the dancers have to undertake rounds of smaller and larger radii. The dancers come together while taking the rounds of smaller radius (near the centre). In this case, the moment of inertia of their system becomes minimum and the frequency increases, to make it thrilling. While outer rounds, the dancers outstretch their legs and arms. This increases their moment of inertia that reduces the angular speed and hence the linear speed. This is essential to prevent slipping.

(2) Diving : Take-off from a springboard or diving platform determines the diver’s trajectory and the magnitude of angular momentum. A diver must generate angular momentum at take-off by moving the position of the arms and by a slight hollowing of the back. This allows the diver to change angular speeds for twists and somersaults in flight by controlling her/his moment of inertia. A compact tucked shape of the body lowers the moment of inertia for rotation of smaller radius and increased angular speed. The opening of the body for the vertical entry into water does not stop the rotation, but merely slows it down. The angular momentum remains constant throughout the flight.

The objects like a cylinder, sphere, wheels, etc. are quite often seen to perform rolling motion. In the case of pure rolling, two motions are undertaking simultaneously; circular motion and linear motion.

Individual motion of the particles (except the one at the centre of mass) is too difficult to describe. However, for theory considerations we can consider the actual motion to be the result of

(i) circular motion of the body as a whole, about its own symmetric axis and

(ii) linear motion of the body assuming it to  be concentrated at its centre of mass. In other words, the centre of mass performs purely translational motion.

Accordingly, the object possesses two types of kinetic energies, rotational and translational. Sum of these two is its total kinetic energy.

E = E_tran + E_rot ... (1)

where E_tran and E_rot are the kinetic energies associated with translation of the center of mass CM and rotation about an axis through the CM, respectively.

Let M and R be the mass and radius of the body.

Let ω, k and I be the angular speed, radius of gyration and moment of inertia for rotation about an axis through its centre, and v be the translational speed of the centre of mass.

v = ωR and I = MK²

E_tran =  \(\frac{1}{2}mv^2\)  and  E_rot = \(\frac{1}{2}Iω^2\)

Total kinetic energy of rolling = Translational K.E. + Rotational K.E.

∴ E = \(\frac{1}{2}Mv^2\) + \(\frac{1}{2}Iω^2\)

= \(\frac{1}{2}Mv^2\) + \(\frac{1}{2}I\frac{v^2}{r^2}\)

= \(\frac{1}{2}Mv^2\)(1+ \(\frac{I}{Mr^2})\)

= \(\frac{1}{2}Mv^2\)(1+ \(\frac{MK^2}{Mr^2})\)

= \(\frac{1}{2}Mv^2\)(1+ \(\frac{K^2}{r^2})\)

Consider a symmetric rigid body, like a sphere or a wheel or a disc, rolling on a plane surface with friction down a plane inclined at an angle q to the horizontal. If the frictional force on the body is large enough, the body rolls without slipping.

Let M and R be the mass and radius of the body.

Let I be the moment of inertia of the body for rotation about an axis through its centre. Let the body start from rest at the top of the incline at a height ft. Let v be the translational speed of the centre of mass at the bottom of the incline. Then, its kinetic energy at the bottom of the incline is

E = \(\frac{1}{2}Mv^2\)(1+ \(\frac{I}{Mr^2}\)

= \(\frac{1}{2}Mv^2\)(1+ β)\)  .... (....β=\(\frac{I}{MR^2}\))

If K is the radius of gyration of the body,

I= MK²  and β=\(\frac{1}{MR^2}\) = \(\frac{K^2}{R^2}\)

From conservation of energy,

(KE + PE)initial = (KE + PE)final

∴ 0 + Mgh =\(\frac{1}{2}Mv^2\)(1+ β)\)+ 0

∴  Mgh =\(\frac{1}{2}Mv^2\)(1+ β)\)

\(v^2=\frac{2gh}{1+β}\)

\(v=\sqrt{\frac{2gh}{1+β}}\)

\(v=\sqrt{\frac{2gh}{1+\frac{k^2}{R^2}}}\)

Since h=Lsinθ

\(v=\sqrt{\frac{2gLsinθ}{1+\frac{k^2}{R^2}}}\)

Let a be the acceleration of the centre of mass of the body along the inclined plane. Since the body starts from rest,

v² = 2aL  ∴ a= v²/2L 

a = \(\frac{2gLsinθ}{1+\frac{k^2}{R^2}}.\frac{1}{2L}\)

= \(\frac{gsinθ}{1+\frac{k^2}{R^2}}\) = \(\frac{gsinθ}{1+β}\)

Starting from rest, if tis the time taken to travel the distance L,

L= (½)at²

∴ t² =\(\sqrt{\frac{2L}{a}}\) = \(\sqrt{\frac{2L}{gsinθ}(1+\frac{K^2}{R^2}}\)

Given- r=0.5 m, ω₀ = 0, ω = 2 rev/s  t = 10 s

(i) Angular acceleration α is constant,

Average angular speed ω_av = \(\frac{ω_0 +ω}{2}\) = \(\frac{0+2}{2}\) = 1 rev/s

∴ Angular displacement of the wheel in time t,

θ = ω_av,×t = 1 × 10 = 10 revolutions

Answer is 10 revolutions

(ii) α = \(\frac{ω-ω_0}{t} = \frac{2-0}{10}=\frac{1}{5}\) rev/s²

θ = ω₀,×t + \(\frac{1}{2} αt^2\) = 0 + \(\frac{1}{2} αt^2\)  (...ω₀ = 0)

∴ for θ₁ = 1 rev

1= \(\frac{1}{2}×\frac{1}{5}t{_1}^2\)

t₁² =10

t₁ = \(\sqrt{10}\)s = 3.162 s

Time for first revolution = 3.162 s

for θ₂ = 9 rev

9= \(\frac{1}{2}×\frac{1}{5}t{_2}^2\)

t₂² =90

∴ t₂ = \(\sqrt{90}\) s = 9.486 s

∴Time for last i.e 10 th revolution is

t - t₂ = 10- 9.486 = 0.514 s

∴Time for first revolution = 0.514 s

Given: m_s=0.5, r₁ = π cm = π x10^-2 m, r₂ = 8 cm = 8x 10^-2 m, g=π² m/s²

To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.

∴ mω²r ≤ μ_s mg

∴ ω²r ≤ μ_s g

4π²f²_min r = μ_s g     …(1)  (…ω=2πf)

∴ for r = r₁

4π²f²_min-1 r₁₌ μ_s g

f²_min1= \(\frac{μ_sg}{4π^2r_1}\) = \(\frac{0.5π^2}{4π^2(π×10^{-2})}\)

f²_min1= \(\frac{100}{8π}=\frac{25}{2π}\)

∴ f_min1 = \(\frac{5}{\sqrt{2π}}\)

At frequency \(\frac{5}{\sqrt{2π}}\) rps, the coin will start slipping

From Eq. 1

f²_min α \(\frac{1}{r}\)

∴ \(\frac{f_{min2}}{f_{min1}}\) =\(\sqrt{\frac{r_2}{r_1}}\) = \(\sqrt{\frac{π}{8}}\)

∴\(f_{min2}\) = \(\sqrt{\frac{π}{8}}\) \(f_{min1}\)

In the second case minimum frequency will be \(\sqrt{\frac{π}{8}}\) times that in the first case

(in text book answer is given for r₁=2cm)

Given : r=72m, v₀ =216 kmph = 216 ×1000/3600 m/s = 60m/s

w= 10m, g= 10m/s²

tanθ = \(\frac{v^2_0}{rg}\) =\(\frac{60^2}{72×10}\)

∴ θ= tan^-15 = 78^°4’

78^°4’ is the required angle of banking.

sinθ=h/w

∴ h =w sinθ = 10 ´ sin78^°4’ = 10 ´ 0.9805

∴ h = 9.8 m

9.8 m is the height of the outer edge of the rack relative to the inner edge.

r=72m, θ= 78°4’(from above question) , u_s=0.8, g= 10m/s² , tanθ= tan78^°4’ =5

v_min = \(\sqrt{rg(\frac{tanθ-u_s}{1+u_stanθ})}\)

= \(\sqrt{72×10(\frac{5-0.8}{1+(0.8×5)})}\) 

v_min = 24.588 m/s = 88.52 km/h

88.52 km/h will be the lower limit or minimum speed on this track.

Since the track is heavily banked q>45° there is no upper limit or maximum speed on this track.

Given : r=6.05m, μ_s = 0.5, g=10m/s² , m=50kg, ∆v=20%

v_min = \(\sqrt{\frac{rg}{μ_s}}\)

∴ v_min = \(\sqrt{\frac{6.05×10}{0.5}} = \sqrt{121}\) =11 m/s

11m/s is the required minimum speed

So long as the cyclist is not sliding, at every instant, the force of static friction is

f_s=mg = 50×10 = 500 N

Given : r=72m, v₀ =216 kmph = 216 ×1000/3600 m/s = 60m/s

w= 10m, g= 10m/s²

tanθ =\(\frac{{v_0}^2}{rg} =\frac{60^2}{72×10}\) = 5

∴θ= tan^-15 = 78^°4’

78^°4’ is the required angle of banking.

sinθ=h/w

∴ h =w sinθ = 10 × sin78^°4’ = 10 × 0.9805

∴ h = 9.8 m

9.8 m is the height of the outer edge of the rack relative to the inner edge.

Given : r=72m,  θ= 78°4’ , μ_s=0.8, g= 10m/s² , tanθ= tan78^°4’ =5

v_min =  \(\sqrt{rg(\frac{tanθ-μ_s}{1+μ_stanθ})}\)

v_min  = \(\sqrt{72×10(\frac{5-0.8}{1+(0.8((5)})}\)

v_min = 24.588 m/s = 88.52 km/h

88.52 km/h will be the lower limit or minimum speed on this track.

Since the track is heavily banked θ > 45° there is no upper limit or maximum speed on this track.

Given : r = 6.05 m, μ_s = 0.5, g = 10 m/s² , m = 50kg, ∆v = 20%

v_min = \(\sqrt{\frac{rg}{μ_s}}\)

= \(\sqrt{\frac{6.05×10}{0.5}}\)

= \(\sqrt{12.1×10} =\sqrt{121}\) = 11 m/s

∴ 11 m/s is the required minimum speed

So long as the cyclist is not sliding, at every instant, the force of static friction is

f_s= mg = 50 × 10 = 500 N

Given : L= 0.2 m, m = 0.1 kg, n = 75 rpm = 75/60 = 5/4 rps, g = 10 m/s² , π² = 10

T=1/n =4/5 s = 0.8 s

T = \(2π\sqrt{\frac{L\,\,cosθ}{g}}\) 

T² = 4π² \(\frac{L\,\,cosθ}{g}\) 

h= L cosθ = \(\frac{gT^2}{4π^2}\)

= \(\frac{10×0.8^2}{4×10}\) =0.16 m  ……(1)

∴ cosθ = 0.16/0.2 =0.8

∴ θ = cos^-1 0.8 =36.87° = 36°5’

v² = rg tanθ = (L sin36.87°)(g)(tan36.87°)     .(....r = Lsinθ)

   =(0.2 × 0.6)(10)(0.75)  = 0.9

The KE of the bob = \(\frac{1}{2}mv^2\) = \(\frac{1}{2}(0.1 × 0.9)\) =  0.045 J

The increase in gravitational PE

∆PE= mg(L− h) =(0.1)(10)(0.2 − 0.16) = 0.04  J

Given : Speed at Top  v_t =6 m/s, Speed at Bottom = v_b =10m/s, g=10 m/s²

v_b²= v_t² + 4gr

∴ r= (v_b² - v_t²)/4g = (10² -6²)/(4´10) = 64/40

∴ r = 1.6m

∴ The diameter of the sphere of death = 3.2m

For this r minimum speed at top

v_min =\(\sqrt{gr}\)  at the top

∴ v_min =\(\sqrt{10×1.6}\) =\(\sqrt{16}\)=4 m/s

Corresponding minimum speed at bottom= \(\sqrt{5gr}\)=\(4\sqrt{5}\)m/s

The required minimum values of the speed are 4 m/s and \(4\sqrt{5}\)m/s  m/s

The MI of the thin ring about its diameter,

I_r = \(\frac{1}{2}\)MR² = 1 kg-m²

Since the ring is melted and recast into a thin disc of same radius R, the mass of the disc equals the mass of the ring = M.

The MI of the thin disc about its own axis (i.e. transverse symmetry axis) is

I_d= \(\frac{1}{2}\)MR² = I_r

∴ I_d = 1 kg-m²

Given : Mass of sphere M_s=50g, Radius of sphere R_s = 10 cm, Mass of rod M_r = 60g, Length of rod L_r = 20 cm

The MI of a solid sphere about its diameter is

I_s-CM = \(\frac{2}{5}\) M_sR_s²

The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere,

h =30 cm.

The MI of a solid sphere about the rotation axis,

I_s  = I_s-CM + M_s h²

For the rod, the rotation axis is its transverse symmetry axis through CM.

The MI of a rod about this axis,

Ir =\(\frac{1}{12}\) M_rL_r²

Since there are two solid spheres, the MI of the dumbbell about the rotation axis is

I= 2I_s + I_r

= 2M_s ( \(\frac{2}{5}\)R_s² + h²) + \(\frac{1}{12}\)M_rL_r²

= 2(50) ( \(\frac{2}{5}\)(10)² + (30)² ) + \(\frac{1}{12}\)(60)(20)²

= 100 (40 + 900) + 5(400) = 94000 + 2000

= 96000 g-cm²

Given : f₁ = 100 rpm, f₂= 80 rpm, M =10 kg, R = 0.4m, m = 1.6kg

I₁=I_wheel = \(\frac{1}{2}MR^2\)= (1/2)(10)(0.4)² = 0.8 kg-m²

MI of the wheel and lump of the clay is

I₂ = I_wheel + mx²

By the principle of conservation of angular momentum

I₁w₁ = I₂w₂

I₁(2πf₁) = I₂(2πf₂)

I₂ = I_wheel + mx² = \(\frac{f_1}{f_2}I_1\) = \(\frac{f_1}{f_2}I_{wheel}\)

∴ mx² = \((\frac{f_1}{f_2}-1)I_{wheel}\)

∴ mx²= \((\frac{100}{80}-1)(0.8)\)= 0.2 kg-m²

∴ x² = 0.2/1.6= 1/8

∴ x=  = 0.3536 m

Given : sinq=3/5, u=0, v = \(\sqrt{10}\) m/s L=5/3 m, g= 10 m/s²

\(v=\sqrt{\frac{2gLsinθ}{1+\frac{k^2}{R^2}}}\)

= \(\sqrt{\frac{2gLsinθ}{1+β}}\)

∴ \(v^2=\frac{2gLsinθ}{1+β}\)

∴ 1+β =\(\frac{2gLsinθ}{v^2}\)  = \(\frac{2×10×(5/2)(2/5}{(\sqrt{10})^2}\) = 2

∴ β = 1 = k²/R²

Therefore the body rolling down is either cylindrical shell or ring