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

AI Thread Summary
The moment of inertia for a solid sphere with a diameter of 0.15m and mass of 0.5kg is calculated as 0.001125 kg*m^2. When the sphere rolls down a ramp and reaches the table, the angular speed is determined to be approximately 44.68 rad/sec after considering both linear and rotational kinetic energy. The initial velocity of the ball upon leaving the table is recalculated to be 5.147 m/s. The distance traveled by the ball after leaving the table is found to be 2.75 meters. The discussion emphasizes the importance of accounting for both types of motion in energy calculations.
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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:
 
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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
 
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?
 
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}
 
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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?
 
can someone please check my work?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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