What is the angular speed of the rotating wheel?

In summary: It should be alpha = r*angular vel. You also need to take into account the initial angular velocity (which is 0 in this case) and the time (2.0s) to solve for the final angular velocity. Overall, your approach is correct.
  • #1
chamonix
21
0

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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  • #2
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.
 
  • #3
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 [tex]\tau[/tex]=I[tex]\alpha[/tex] , I calculated 0.0958

c) I've calculated 14.1 m/s from using a=r[tex]\alpha[/tex]
then I used [tex]\alpha[/tex]=r[tex]\omega[/tex]2 and solved from there. does that sound correct?
 
  • #4
sorry for part b I got I=0.0958
 
  • #5
anyone to confirm?
 
  • #6
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
 
  • #7
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.
 

1. What is rotational equilibrium?

Rotational equilibrium is a state in which an object is not rotating or is rotating at a constant rate without any external forces causing it to accelerate or decelerate.

2. How is rotational equilibrium different from translational equilibrium?

Rotational equilibrium refers to a balance of forces and torques acting on an object, while translational equilibrium refers to a balance of forces only. In rotational equilibrium, the object may still be moving, but its rotational motion remains constant.

3. What are the conditions for rotational equilibrium?

The two conditions for rotational equilibrium are that the net torque acting on the object must be zero and the center of mass must not accelerate. This means that the sum of all the external forces acting on the object must be zero and the sum of all the torques acting on the object must also be zero.

4. How is torque calculated in rotational equilibrium?

Torque is calculated by multiplying the force applied to an object by the distance from the pivot point, or axis of rotation. It is represented by the equation τ = r x F, where τ is torque, r is the distance from the pivot point, and F is the applied force.

5. How is rotational equilibrium important in everyday life?

Rotational equilibrium is important in everyday life because it allows objects to remain stable and maintain their position without falling or tipping over. It is also crucial in many mechanical systems, such as gears and pulleys, which rely on rotational equilibrium to function properly.

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