A uniform solid sphere rolls down a vertical surface without sliding. If the vertical surface moves with an acceleration a = g 2 , find the minimum coefficient of friction between the sphere and vertical surfaces so as to prevent relative sliding.

We have to find out the minimum coefficient of friction between the sphere and vertical surfaces to prevent relative sliding motion. In the diagram α is angular acceleration in a clockwise direction. Now if we consider horizontal equilibrium then Normal force N = m g 2 . . . ( 1 ) And for vertical motion equation would be m g - f = m a ' . . . ( 2 ) And torque equation would be f × R = 2 5 m R 2 × α . . . ( 3 ) Pure rolling is given by a ' = α R so ( 3 ) becomes f × R = 2 5 m a ' R friction f = 2 5 m a ' . . . ( 4 ) so ( 2 ) becomes m g - 2 5 m a ' = m a ' m g = 2 5 + 1 m a ' m g = 7 5 m a ' a ' = 5 7 g so ( 4 ) becomes f = 2 5 × m × 5 7 × g f = 2 7 m g So the coefficient of friction is the ratio of the friction force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. μ = F N Where F is the frictional force and N is the normal force. μ = 2 7 m g × 2 m g μ = 4 7 This is the expression for coefficient of friction for no slipping.

A uniform solid sphere rolls down a vertical surface without sliding. If the vertical surface moves with an acceleration a = g 2 , find the minimum coefficient of friction between the sphere and vertical surfaces so as to prevent relative sliding.

We have to find out the minimum coefficient of friction between the sphere and vertical surfaces to prevent relative sliding motion. In the diagram α is angular acceleration in a clockwise direction. Now if we consider horizontal equilibrium then Normal force N = m g 2 . . . ( 1 ) And for vertical motion equation would be m g - f = m a ' . . . ( 2 ) And torque equation would be f × R = 2 5 m R 2 × α . . . ( 3 ) Pure rolling is given by a ' = α R so ( 3 ) becomes f × R = 2 5 m a ' R friction f = 2 5 m a ' . . . ( 4 ) so ( 2 ) becomes m g - 2 5 m a ' = m a ' m g = 2 5 + 1 m a ' m g = 7 5 m a ' a ' = 5 7 g so ( 4 ) becomes f = 2 5 × m × 5 7 × g f = 2 7 m g So the coefficient of friction is the ratio of the friction force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. μ = F N Where F is the frictional force and N is the normal force. μ = 2 7 m g × 2 m g μ = 4 7 This is the expression for coefficient of friction for no slipping.

A rod of mass m spins with an angular speed ω = g / l , Find its (a) kinetic energy of rotation. (b) kinetic energy of translation (c) total kinetic energy.

(a) Kinetic energy of rotation can be calculated by formula K rotational = 1 2 I C ω 2 where I C is the moment of inertia of centre of mass which is equal to m l 2 12 for the rod and ω is the angular speed of rotation. ⇒ K rotational = 1 2 m l 2 12 g l 2 ⇒ K r o t a t i o n a l = 1 24 m g l (b) Kinetic energy of translation can be calculated by formula K translational = 1 2 m v c 2 where v c is the velocity of centre of mass of rod which can be calculated by formula v c = ω l 2 . ⇒ K translational = 1 2 m ω l 2 2 ⇒ K translational = 1 2 m g l × l 2 2 ⇒ K translational = 1 8 m g l (c) We know that total kinetic energy is equal to translational kinetic energy and rotational kinetic energy. Hence, we can write K total = K translational + K rotational ⇒ K total = 1 2 I P ω 2 + 1 2 m l 2 3 g l 2 ⇒ K t o t a l = 1 6 m g l

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A uniform solid sphere rolls down a vertical surface without sliding. If the vertical surface moves with an acceleration $a = \frac{g}{2}$ , find the minimum coefficient of friction between the sphere and vertical surfaces so as to prevent relative sliding.

Important Questions on Rigid Body Dynamics: Part 2

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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 5

A spool (consider it as a double disc system joined by a short tube at their centre) is placed on a horizontal surface as shown in figure. A light string wound several times over the short connecting tube leaves it tangentially and passes over a light pulley. A weight of mass m is attached to the end of the string. The radius of the connecting tube is r and mass of the spool is $M$ and radius is $R$ . Find the acceleration of the falling mass $m$ . Neglect the mass of the connecting tube and slipping of the spool.
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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 6

A solid cylinder wheel of mass

M

and radius

f = \frac{2}{5} \times m \times \frac{5}{7} \times g

is pulled by a force

F

applied to the centre of the wheel at

37 °

to the horizontal. If the wheel is to roll without slipping, what is the maximum value of

|F|

? The coefficients of static and kinetic friction are

μ_{S} = 0.40

and

μ_{K} = 0.30 . (\sin 37 ° = \frac{3}{5})

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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 7

A uniform solid cylinder of mass $m$ , rests on two horizontal planks. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically down, with a constant force $F$ . Find the maximum magnitude of the force $F$ , which still does not bring about any sliding of the cylinder, if the coefficient of friction between the cylinder and the planks is equal to $k$ . What is the acceleration $ω_{\max}$ , of the axis of the cylinder, rolling down the inclined plane?

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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 9

A solid sphere of mass $M$ is placed on the top of a plank of the same mass, after giving an angular velocity $ω_{0}$ at $t = 0$ . Find the velocity of the plank and the sphere when the sphere starts rolling.
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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 1

A rod of mass $m$ spins with an angular speed $ω = \sqrt{g / l}$ , Find its
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(a) kinetic energy of rotation.
(b) kinetic energy of translation
(c) total kinetic energy.

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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 2

Find $K E$ of the rotating composite rod comprising two rods of mass $m$ and $2 m$ and each of length $l = 1 m$ . Assume $ω = 10$ rad $s^{- 1}$ and $m = 1 k g$ .
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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 3

A uniform rod smoothly pivoted at one of its ends is released from rest. If it swings in vertical plane, find the, maximum speed of the end $P$ of
the rod.
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Physics For Joint Entrance Examination JEE (Advanced): Mechanics II>Chapter 4 - Rigid Body Dynamics: Part 2>CONCEPT APPLICATION EXERCISE>Q 4

A rigid body is made of three identical uniform thin rods each of length $L$ fastened together in the form of letter $H .$ The body is free to rotate about a fixed horizontal axis $AB$ that passes through one of the legs of the $H .$ The body is allowed to fall from rest from a position in which the plane of $H$ is horizontal. What is the angular speed of the body, when the plane of $H$ is vertical.

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A uniform solid sphere rolls down a vertical surface without sliding. If the vertical surface moves with an acceleration a=g2, find the minimum coefficient of friction between the sphere and vertical surfaces so as to prevent relative sliding.

Important Questions on Rigid Body Dynamics: Part 2

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A uniform solid sphere rolls down a vertical surface without sliding. If the vertical surface moves with an acceleration $a = \frac{g}{2}$ , find the minimum coefficient of friction between the sphere and vertical surfaces so as to prevent relative sliding.