A solid spherical ball with diameter of 2 cm rolls down a ramp without slipping.
ID: 1500744 • Letter: A
Question
A solid spherical ball with diameter of 2 cm rolls down a ramp without slipping. The distance from the base of the ramp to the initial location of the disk is 1.8 m and the ramp is inclined at an angle of 30 degrees above the horizontal. a. Find the moment of inertia (i.e. the rotational inertia, I) of the sphere/ball. b. If the disk is released from rest what will its total kinetic energy when it reaches the bottom of the ramp? c. What will be its speed at the bottom of the ramp? d. What will its translational kinetic energy be at the bottom? What about the rotational kinetic energy?
Explanation / Answer
given
d = 2 cm
r = 2/2 = 1 cm = 0.01 m
vertical height of the ramp, h = d*sin(30)
= 1.8*sin(30)
= 0.9 m
let m is the mass of the ball.
a) I = (2/5)*m*r^2
b) final total kinetic energy = Initial potentail energy
= m*g*h
c) Apply conservation of energy
final kinetic energy = initial potentail energy
0.5*m*v^2 + 0.5*I*w^2 = m*g*h
0.5*m*v^2 + 0.5*(2/5)*m*r^2 = m*g*h
0.5*m*v^2 + 0.2*m*v^2 = m*g*h
0.7*m*v^2 = m*g*h
v = sqrt(g*h/0.7)
= sqrt(9.8*0.98/0.7)
= 3.7 m/s
d) KE_translational = 0.5*m*v^2
= 0.5*m*3.7^2
e) KE_rotational = 0.2*m*v^2
= 0.2*m*3.14^2
we need m value to find ll these values
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