A solid ball with a mass of 200 g and a radius of 2.5 cm is placed on a friction

Question

A solid ball with a mass of 200 g and a radius of 2.5 cm is placed on a frictionless ran inclined at 20 degree and released from a height of 30 cm. It slides without rolling down the Once it reaches the floor, a frictional force of 50 N acts on the ball until it rolls without slipping. 1. What is the speed of the ball when it reaches the floor? 2. What is the acceleration of the ball as it slides along the floor? 3. What is the torque on the ball as it slides along the floor? 4. What is the angular acceleration of the ball as it slides along the floor? 5. If t = 0 when the ball reaches the floor, at what time will v = R_ omega ? 6. What will the speed of the ball be when it rolls without slipping? 7. How much kinetic energy will be dissipated as the ball slides? A red cart with a mass of 250 g is initially moving east on a track at 1.25 m/s. It collides a green cart with a mass of 400 g that is initially moving west on the track at 0.50 m/s. During the collision, 0.15 J are dissipated. 8. What are the total momentum and kinetic energy before the collision? 9. How much momentum is lost during the collision? 10. What are the final velocities of each cart? 11. If the collision point is at x=0 and east is +x, where is the center of mass of the t cart system at t=2.0 s?

Explanation / Answer

soving first 4 parts of the question

(1)We know moment of inertial of solid sphere =I=(2/5)MR^2=5*10^(-5)

Conserving Energy

let velocity is v at the bottom

mgh=0.5mv^2+0.5*I*w^2

if no sipping ,then

w=v/R

mgh=0.5mv^2+0.5*I*(v/R)^2

we got v=2.05 m/s

(2)let accleration is a m/s^2

frictional force=0.5

a=-0.5/m=-2.5 m/s^2

(3)torque=frictional force*radius=0.0125 Nm

(4)angular accleration=a/R=100 rad/s^2

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