Physics Help

[buffgilville]

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.

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! :smile:

[Pyrrhus]

Remember when both movement (linear and rotational) are combined

K = \frac{1}{2}mv^{2}_{cm} + \frac{1}{2}I_{cm} \omega^2

You can rearrange this to find \omega

remember v = \omega r

K = \frac{1}{2}m \omega^{2} r^{2} + \frac{1}{2}I_{cm} \omega^2

[buffgilville]

ok, so
b) w = 44.92 rad/sec
c) Would the initial velocity of the ball when it leaves the table still be 3.987m/s?

[Pyrrhus]

You only considered linear movement kinetic energy, you have an object with linear and rotational motion. Refer to my post for \Delta K You can express the direction by using \pm v_{x} \hat{i}

[buffgilville]

Ok, I redid the problem considering, like you said, the kinetic energy of both linear and rotational movement.

so,
(0.5)(9.81)(0.81) = (1/2)(0.5)(0.075^2)(w^2) + (1/2)(0.001125)w^2 ---> w=44.68 rad/sec.

and velocity when the ball leaves the table is 5.147m/s

and d = 2.75meters

right?

[buffgilville]

can someone please check my work?