A solid sphere of diameter 0.15m and mass 0.5kg is released and rolls without slipping down the ramp. The ramp is 0.81m high and is on a table. The table has a height of 1.40 meters. The balls move freely once it leaves the table.(adsbygoogle = window.adsbygoogle || []).push({});

a) What is the moment of inertia of the ball about the axis of rotation?

I = (2/5)(0.5kg)(0.075)^2 =0.001125 kg*m^2

b) Find the angular speed of the ball when it reaches the table top.

Potential energy = Kinetic Energy

(0.5)(9.81)(0.81) = (1/2)(0.001125)w^2 --->w=84.04 rad/sec.

c) Find the initial velocity (magnitude and direction) of the ball when it leaves the table.

Total Energy = 0 = (1/2)(0.5)(v)^2 - (1/2)(0.001125)(84.04)^2

velocity = 3.987m/s

How do I find the direction?

d) Find the distance d (distance ball travels when it left the table).

x = vt + (1/2)at^2

-1.40y + dx = (3.987x)t + (-4.905y)t^2

x: d = 3.987t

y: -1.40 = -4.905t^2 ----> t=0.534sec.

d=3.987(0.534) ---->d=2.13meters

Are my answers correct? Did I do anything wrong? Please help. Thanks!

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