A tennis ball is rolling up a ramp It has a radius of 34 cm

A tennis ball is rolling up a ramp. It has a radius of 3.4 cm, and an initial angular speed of 18 rad/s. (a) What is the ball\'s translational speed when its angular speed is 18 rad/s? (b) The ball makes exactly 13.0 revolutions while rolling uphill to a stop. Calculate its angular displacement in radians. (c) Calculate the ball\'s angular acceleration while slowing down.

Solution

a.) For a ball moving with an angular velocity of w rad/s, the linear velocity is given as: v = wR m/s

Where R is the radius of the ball.

Hence for the given situation, we have w = 18 rad/s and R = 3.4 cm

Therefore v = 18 x 3.4 = 61.2 cm/s

b.) For the 13 revolutions along the ramp, the distance covered would be equal to 13 times the circumference of the ball.

That is, distance = 13 x 2 x 3.14 = 81.64 radians

c.) For the motion along the ramp we can write that:

v² = u² - 2aS

or, 0 = 324 - 2a x 81.64

or, a = 1.9843 rad/s² is the required angular deceleration.

d.) As the angular acceleration is give as Torque / Inertia

We can write: Torque / Inetia = 1.9843

or, Inertia = 9 x 10^-5 / 1.9843 = 4.5356 x 10^-5 Kg-m²

7.) For an ice skater moving spinning around, we do not have any external force acting on him, hence the angular momentum remains conserved. Also, on stretching the arms we know that the mass gets spread over a larger distance from the axis of rotation hence increasing the moment inertia. We know that for a given inertia, momentum is directly proportional to the angular velocity. Therefore, on increasing the inertia, the angular velocity decreases for the given angular momentum. That is why the skater slows down on spreading the arms.