Name: Kinetic Energy of Rotational Motion
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: We all know the formula for translational kinetic energy, but we need rotational kinetic energy for a rotating object. Let's learn the expression for it.

Question

A bowling ball of mass

M

and radius

R

is thrown so that the instant it touches the floor it is moving with a velocity

5 m / s

horizontally and is not rotating. The coefficient of friction between the ball and the floor is

0.08

. Find the time the ball slides.

Accepted Answer

1.82 s

Given the mass and radius of the ball= $M and R$

The velocity of the ball $v = 5 m / s$

The Coefficient of friction= $0.08$

When the ball is slipping, the forces acting on it are:
1) Weight $Mg$ , acting vertically downward through the centre,
2) Normal reaction $N$ , vertically upward through the point of contact
3) Forces of kinetic friction, $f_{k} = μN$ , horizontally through the point of contact and opposite to the direction of motion.The horizontal acceleration produced by this force.

$a = \frac{f_{k}}{M} = μg$ as $N = mg$ ( directed opposite to the direction of motion.

Therefore, its velocity at any time $t$ is given by
$V = V_{0} - μgt$ .........................................(1)
Further, the only force that produces a torque about the centre is $f_{k}$ This torque is given by
$τ = f_{k} R$ , acting in an anticlockwise direction.

The angular acceleration produced by this torque is $α$ . The acceleration, α of the ball about its centre given as $f_{k} R = I_{cm} . α or μgMgR = \frac{2}{5} {MR}^{2} α or α = \frac{5 μg}{2 R}$

This angular acceleration sets the ball rotating with increasing angular velocity in anticlockwise direction whose magnitude $ω$ , at any instant $t$ is given by
$ω = αt$ .........................................................(2)
When the ball starts pure rolling,
$V = ωR$ ..........................................................(3)
Hence, using equations (1) , (2) and (3) we get,

$V_{0} - μgt = αtR = \frac{5 μgt}{2} or μgt = \frac{2 v_{0}}{7} or t = \frac{2 V_{0}}{7 μg}$

Substituting the given values, we have,

The time the ball slides $t = \frac{2 \times 5}{7 \times 0.08 \times 9.8} = 3.55 s e c .$

A bowling ball of mass $M$ and radius $R$ is thrown so that the instant it touches the floor it is moving with a velocity $5 m / s$ horizontally and is not rotating. The coefficient of friction between the ball and the floor is $0.08$ . Find the time the ball slides.

Important Questions on Motion of System of Particles and Rigid Body

Learn and Prepare for any exam you want !

A bowling ball of mass M and radius R is thrown so that the instant it touches the floor it is moving with a velocity 5 m/s horizontally and is not rotating. The coefficient of friction between the ball and the floor is 0.08. Find the time the ball slides.

Important Questions on Motion of System of Particles and Rigid Body

Learn and Prepare for any exam you want !

A bowling ball of mass $M$ and radius $R$ is thrown so that the instant it touches the floor it is moving with a velocity $5 m / s$ horizontally and is not rotating. The coefficient of friction between the ball and the floor is $0.08$ . Find the time the ball slides.