Question

A 1.0 kg spherical mass with a 10-cm radius rolls without slipping 5.0 meters down a 30degree incline. Isphere=(2/5)MR^2

a. what is the angular velocity of the sphere at the bottom of the ramp?

b. through what angle has the sphere rotated?

c. what is the angular acceleration of the sphere as it rolls down the incline?

d. how much time did it take to roll down the incline?

Answer 1

for solid sphere,

let v is the linear speed and w is the angular speed of sphere at the bottom of the ramp.
Apply conservation of energy

final kinetic energy = initial potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/5)m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*v^2 = m*g*h

(7/10)m*v^2 = m*g*h

v = sqrt(10*g*h/7)

= sqrt(10*9.8*5*sin(30)/7)

= 5.92 m/s

w = v/r = 5.92/0.1 = 59.2 rad/s <<<<<<-----------Answer

b) angular displacement of the sphere, theta = L/r = 5/0.1 = 50 radians <<<<<<-----------Answer

c) alfa = (wf^2 - wi^2)/(2*theta)

= (59.2^2 - 0^2)/(2*50)

= 35.0 rad/s^2 <<<<<<-----------Answer

d) t = (wf - wi)/alfa

= (59.2 - 0)/35

= 1.69 s <<<<<<-----------Answer