Question

From rest a ball of radius 6 cm rolls down a 1.2 m high hill. Using conservation of energy calculate the angular speed of the ball at the bottom. (I =2mr2/5)

Answer 1

Using energy conservation on the ball at initial and final position

KEi + PEi = KEf + PEf

PEi = gravitational potential energy = 0, at ground

PEf = m*g*h

KEi = 0, since ball starts from rest

KE = KEtrans + KErot

KEf = 0.5*m*V^2 + 0.5*I*w^2

I = Moment of inertia of solid sphere = 2*m*R^2/5

w = final angular speed at the bottom = V/R

Using these values:

0 + m*g*h = (1/2)*m*V^2 + (1/2)*(2/5)*m*R^2*(V/R)^2 + 0

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

g*h = 7*V^2/10

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

given that h = height of hill = 1.2 m

So,

V = sqrt (10*9.81*1.2/7)

V = final linear speed of ball at the bottom = 4.10 m/s

w = V/R

R = radius of ball = 6 cm = 0.06 m

So,

w = V/R = 4.10/0.06

w = final angular speed = 68.33 rad/sec

Let me know if you've any query.