What is the angular speed of the rotating wheel?

AI Thread Summary
The discussion revolves around calculating the angular speed of a rotating wheel attached to a mass on an incline. Participants are attempting to solve for the force in the rope, the moment of inertia of the wheel, and the angular speed after 2 seconds. Key equations include F=ma for force calculations and I=1/2mr^2 for moment of inertia, though the mass of the wheel is not provided. The correct approach involves relating linear acceleration to angular acceleration and using rotational kinematics to find angular speed. The conversation highlights the importance of correctly identifying forces and applying the right formulas to solve the problem.
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Homework Statement


A 12 kg mass is attached to a cord that is wrapped around a wheel with a radius of 10.0 cm. The acceleration of the mass down the frictionless incline of 37 degrees is measured to be 2.0 m/s^2. Assuming the axle of the wheel to be frictionless, determine:
a). the force in the rope.
b). the moment of inertia of the wheel.
c). the angular speed of the wheel 2.0 s after it begins rotating, starting from rest.

Homework Equations


I=1/2mr^2



The Attempt at a Solution


For a). I got 14.44 N by using sin37*2*12
for b). I used I=1/2mr^2 and got .06
for c). I did not understand how to get this answer.
Please help. Thank You.
 
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I got 14.44 N by using sin37*2*12
Not quite right. Set up an equation with F=ma, and sum up the forces. Remember, the block is accelerating, it's not at rest.
I used I=1/2mr^2 and got .06
But you don't know the mass of the wheel, 12 kg is the mass of the block. The tension due to the block is causing a torque on the wheel. You must relate the two to find the moment of inertia.
I did not understand how to get this answer.
You were given the linear acceleration so you can get the angular acceleration. From that, using rotational kinematics, you can find the angular speed.
 
does anyone know about this problem?

for a) what are the external forces? I said F=ma = (12.0 kg)(cos 37)(2.0m/s) = 19.17 N

b) I said Torque = radius*force = (0.1 m)(12 kg*cos(37)*2.00m/s) = 1.92 N*m

thus solving for in \tau=I\alpha , I calculated 0.0958

c) I've calculated 14.1 m/s from using a=r\alpha
then I used \alpha=r\omega2 and solved from there. does that sound correct?
 
sorry for part b I got I=0.0958
 
anyone to confirm?
 
Answer:

a)

F= mgsin(theta) - ma

b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha.

c) alpha = r*angular vel^2 , solve for ang vel and your done
 
Hi nopistons93,

nopistons93 said:
Answer:

a)

F= mgsin(theta) - ma

b) Use T=rF and solve for T , then plug T into T=I*alpha and solve for I using a/r to find alpha.

c) alpha = r*angular vel^2 , solve for ang vel and your done

The formula highlighted in red is not correct.
 
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