Starting from rest, a basketball rolls from the top to the bottom of a hill, rea
ID: 1791275 • Letter: S
Question
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.3 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?
When answering please assume I don't understand anything, which is true. I looked at other answers and they gave an equation like "I = 2/3*mr^2 for basketball" and I see we only got one piece of data so I don't understand where "m" or "r" are supposed to come from? and what is I? Please explain why you're using each equation as well. Thank you!!!
Explanation / Answer
(a)
When the velocity of the ball is v its angular velocity is w = v/r
The I for ball (hollo sphere ) is 2/3* m*r2.
The rotational kinetic energy of ball (hollow sphere) = 0.5* I* w2= 0.5* (2/3) *m*r2 * (v/r) 2 = m*v2/3
The linear kinetic energy = 0.5*m*v2
Total kinetic energy at any instant = m*v2/2 + m*v2/3 = (5/6) m*v2
nitially its potential energy = m*g*h
Final total energy = (5/6)* m*v2 since potential energy at the bottom is zero.
(5/6)*m*v2 = m*g(h
h = (5/6)* v2/g
h = (5/6)*(6.3)2 / 9.8
h = 3.37 m
(b)
As before
m*g*h = 0.5*m*v2 + 0.5*I*w2
I for solid cylinder = 0.5*m*r2
. m*g*h = (1/2)*m*v2 + (1/4)*m*r2*(v/r)2
m*g*h = 0.75*m*v2
v2 = 9.8*3.37 /0.75
v = 6.63 m/s
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