How Does Friction Affect Wheel Motion on a Horizontal Surface?

[kara]

The question is:

A wheel is rolling smoothly on horizontal surface. There is a constant horizontal force of magnitude 10 N. Wheel has mass of 10 kg, and radius of 0.3 m. The accel. of its centre of mass has mag. of 0.6 m/s^2.

I am asked to find frictional force in unit-vector notation. This is what i did:
(10N) = f - (10kg)(9.8m/s^2)sin0* = (10kg)(0.6 m/s^2)
(10N)-f = 6.0
-f = -4.0

Since there is clockwise angluar acccel the friction is a negative value. So my answer is f = (-4.00 N)i

For part b it asks for the rotational inertia of the wheel about its centre of mass. This is what I've done so far:

(-4 N)(0.3m) = I (-0.6m/s^2/0.3m)
-1.20 = I(-2)
I = 0.600 rad/s

Am i on the right track?? Please help!

 

[OlderDan]

Friction tends to retard the motion, and I assume the applied force and the acceleration are in one direction, with friction in the opposite direction. Your positive answer for f is because you have it in the wrong place in your equation for it to be negative. If you wrote F + f = ma, then f would be found to be negative. Your result for I has the wrong dimensions.

 

[kyle8921]

Yes, you are on the right track. Your calculation for the frictional force in unit-vector notation is correct. The negative value indicates that the frictional force is acting in the opposite direction of the horizontal force, which makes sense since the wheel is rolling in the direction of the horizontal force.

For part b, you are correct in using the formula I = τ/α to calculate the rotational inertia. However, there are a few errors in your calculation. First, the unit of rotational inertia is kg*m^2, not rad/s. So the correct unit for the left side of the equation would be kg*m^2. Second, the value of α that you used is incorrect. The correct value of α in this case is 0.6 m/s^2, which is the acceleration of the center of mass. Finally, the unit of τ in this equation is N*m, not N. So the correct unit for the right side of the equation would be N*m. Putting all of this together, the correct equation is:

-4 N * 0.3 m = I * (0.6 m/s^2 / 0.3 m)
-1.2 N*m = I * 2 N
I = 0.6 kg*m^2

So the rotational inertia of the wheel about its center of mass is 0.6 kg*m^2. This means that it would take a torque of 0.6 N*m to produce an angular acceleration of 1 rad/s^2.