A uniform hollow sphere of mass m = 1 kg is placed on a rough horizontal surface for which the coefficient of static friction between the surfaces in contact is μ = 2 5 . Find the maximum constant force which can be applied at the highest point in the horizontal direction so that the sphere can roll without slipping. ( g = 10 m s - 2 )

In absence of friction if force F is applied then linear acceleration a and angular acceleration α will be a = F m and α = τ I = F R 2 3 m R 2 = 3 F 2 m R as α R > a ⇒ In absence of friction there will be backward slipping so, friction force f will act along the same direction as that of applied force F . So, a = F + f m . . i and α = F - f R 2 3 m R 2 = 3 F - f 2 m R … ii As for pure rolling a = α R and using i and ii F + f m = 3 F - f 2 m R R ⇒ 2 F + f = 3 F - f ⇒ F = 5 f As maximum value possible for f = μ m g = 2 5 × 1 kg × 10 m s - 2 = 4 N ⇒ The maximum possible value of F = 5 × 4 N = 20 N

A uniform solid sphere is placed on a smooth horizontal surface. An impulse I is given horizontally to the sphere at a height h = 4 R / 5 above the centre line. m and R are mass and radius of the sphere, respectively. a Find the angular velocity of the sphere and linear velocity of centre of mass of the sphere after impulse. b Find the minimum time after which the highest point B will touch the ground, c Find the displacement of the centre of mass during this interval.

a Angular impulse about a point is equal to the product of impulse and the perpendicular distance of application of impulse from the point, mathematically J = I r ⊥ = I h … i Now, angular impulse = change in angular momentum = Moment of inertia × change in angular velocity ⇒ J = 2 5 m R 2 ∆ ω ⇒ I h = 2 m R 2 ∆ ω 5 ⇒ I 4 R 5 = 2 m R 2 ∆ ω 5 ⇒ ∆ ω = 2 I m R . As initial angular velocity is zero ⇒ ω = 2 I m R . Again impulse = change in linear momentum = Mass × change in velocity ⇒ I = m ∆ v ⇒ ∆ v = I m . As initial velocity is zero ⇒ v = I m . b The highest point B will touch the ground when the sphere rotates by an angle π ⇒ ω t = π ⇒ t = π ω = π 2 I m R = m R π 2 I . c As COM is moving horizontally with constant velocity, displacement at the same time t will be S = v t = I m m R π 2 I = π R 2 .

Consider a bicycle in vertical position accelerating forward without slipping on a straight horizontal road. The combined mass of the bicycle and the rider is M and the magnitude of the accelerating torque applied on the rear wheel by the pedal and gear system is τ . The radius and the moment of inertia of each wheel is R and I (with respect to the axis) respectively. The acceleration due to gravity is g . (a) Draw the free diagram of the system (bicycle and rider ). (b) Obtains the acceleration a in terms of the above-mentioned quantities. a = (c) For simplicity assume that the centre of mass of the system is at height R from the ground and equidistant at 2 R from the centre of each of the wheels. Let μ be the coefficient of friction (both static and dynamic) between the wheels and the ground. Consider M > > I R 2 and no slipping. Obtain the conditions for the maximum acceleration a m of the bike. a m = (d) For μ = 1 . 0 calculate a m . a m =

(a) The friction forces f 1 and f 2 on the wheels and normal N 1 and N 2 on the wheels are shown on figure below, Here, for pure rolling a = α R (b) From the free body diagram, we have, N 1 + N 2 = M g . . . 1 f 2 - f 1 = M a . . . 2 In Wheel 2 , a torque τ is applied due to wheel and gear, so, τ - f 2 R = I α = I a R . . . 3 In Wheel 1 , f 1 R = I α = I a R . . . 4 From equation 2 , 3 and 4 , a = τ R M R 2 + 2 I (c) When the acceleration will become maximum, the torque on Wheel 2 will be maximum and the friction on it will also be maximum, i.e., f 2 = μ N 2 , put this in all above equations, μ N 2 - f 1 = M a m . . . 5 τ m - μ N 2 R = I a m R . . . 6 f 1 R = I a m R . . . 7 a m = τ m R M R 2 + 2 I . . . 8 The net torque about the centre of mass will be zero, so, ∑ τ C M = 0 N 2 2 R + f 1 R = N 1 2 R + μ N 2 R 2 N 2 + f 1 = 2 N 1 + μ N 2 . . . 9 From equation 1 , 2 N 2 + f 1 = 2 M g - N 2 + μ N 2 4 N 2 + f 1 = 2 M g + μ N 2 From equation 5 , 4 N 2 = 2 M g + M a m . . . 10 From equation 6 and 10 , τ m = a m M R + 2 I R . . . 11 From equation 8 and 11 , a m = μ g 2 1 - μ 4 (d) For μ = 1 , a m = 1 g 2 1 - 1 4 = 2 g 3 .

HARD

JEE Main/Advance

IMPORTANT

Earn 100

Calculate the moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to m and the radius of its base to $R$ .

Important Questions on Rotational Mechanics

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 1>Q 40

A uniform hollow sphere of mass

m = 1 kg

is placed on a rough horizontal surface for which the coefficient of static friction between the surfaces in contact is

μ = \frac{2}{5}

. Find the maximum constant force which can be applied at the highest point in the horizontal direction so that the sphere can roll without slipping.

(g = 10 m s^{- 2})

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 1>Q 42

A uniform solid sphere is placed on a smooth horizontal surface. An impulse $I$ is given horizontally to the sphere at a height $h = 4 R / 5$ above the centre line. $m$ and $R$ are mass and radius of the sphere, respectively.

Question Image

$(a)$ Find the angular velocity of the sphere and linear velocity of centre of mass of the sphere after impulse.

$(b)$ Find the minimum time after which the highest point $B$ will touch the ground,

$(c)$ Find the displacement of the centre of mass during this interval.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 1>Q 44

A solid cubical block of mass

m

and side

a

slides down a rough inclined plane of inclination

θ

with a constant speed. Calculate the torque of the normal force acting on the block about its centre and the perpendicular distance

x

from centre of mass at which it is acting.

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 1>Q 47

The moment of inertia of a uniform semicircular wire of mass

M

and radius

r

about a line perpendicular to the plane of the wire through the centre is:

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 4>Q 24

A uniform cylinder of radius $R$ and mass $M$ can rotate freely about a stationary horizontal axis $O$ . A thin cord of length $l$ and mass $m$ is wound on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length $x$ of the hanging part of the cord. The wound part of the cord is supposed to have its centre of gravity on the cylinder axis.

Question Image

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 4>Q 26

A point $A$ is located on the rim of a wheel of radius $R = 0.50 m$ which rolls without slipping along a horizontal surface with velocity $v = 1.00 m / s$ as shown in figure. Find:

(a) the modulus and the direction of the acceleration vector of the point $A$ ;

(b) the total distance s traversed by the point $A$ between the two successive moments at which it touches the surface.

Question Image

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 4>Q 27

A uniform sphere of mass $m$ and radius $r$ rolls without sliding over a horizontal plane, rotating about a horizontal axle $OA$ . In the process, the centre of the sphere moves with velocity along a circle of radius $R$ Find the kinetic energy of the sphere.

Question Image

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Physics>Chapter 10 - Rotational Mechanics>Exercise - 4>Q 30

Consider a bicycle in vertical position accelerating forward without slipping on a straight horizontal road. The combined mass of the bicycle and the rider is $M$ and the magnitude of the accelerating torque applied on the rear wheel by the pedal and gear system is $τ$ . The radius and the moment of inertia of each wheel is $R$ and $I$ (with respect to the axis) respectively. The acceleration due to gravity is $g$ .

(a) Draw the free diagram of the system (bicycle and rider ).

Question Image

(b) Obtains the acceleration $a$ in terms of the above-mentioned quantities. $a =$

(c) For simplicity assume that the centre of mass of the system is at height $R$ from the ground and equidistant at $2 R$ from the centre of each of the wheels. Let $μ$ be the coefficient of friction (both static and dynamic) between the wheels and the ground. Consider $M > > \frac{I}{R^{2}}$ and no slipping. Obtain the conditions for the maximum acceleration $a_{m}$ of the bike. $a_{m} =$

(d) For $μ = 1.0$ calculate $a_{m}$ . $a_{m} =$

Calculate the moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to m and the radius of its base to R.

Important Questions on Rotational Mechanics

Learn and Prepare for any exam you want !

Calculate the moment of inertia of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to m and the radius of its base to $R$ .