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

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Homework Help Overview

The discussion revolves around the moment of inertia of a solid sphere and its motion as it rolls down a ramp and leaves a table. The problem involves concepts from dynamics, including energy conservation and rotational motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of the moment of inertia and its implications for the angular speed and linear velocity of the ball. There are questions about the correctness of the initial calculations and the consideration of both linear and rotational kinetic energy.

Discussion Status

Some participants have provided alternative calculations and corrections regarding the angular speed and the initial velocity of the ball. There is ongoing verification of the results and consideration of the combined effects of linear and rotational motion.

Contextual Notes

Participants are discussing the assumptions made in the calculations, particularly regarding the treatment of kinetic energy and the direction of motion. There is a focus on ensuring that both linear and rotational aspects are accounted for in the analysis.

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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:
 
Last edited:
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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}
 
Last edited:
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?
 

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