Calculating Moment of Inertia of Sphere w/ Mass 100kg & Radius 1m

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To calculate the time it takes for a solid sphere with a mass of 100 kg and a radius of 1 m to stop when subjected to a braking force of 1 N, the moment of inertia is first determined using the formula I = (2/5)MR^2, resulting in 40 kg m². The torque is calculated using τ = r x F, yielding τ = 1 N m. The angular acceleration is then found by dividing torque by moment of inertia, giving an angular acceleration of 0.025 rad/s². The time to stop is calculated using the formula t = (w - w0) / α, which results in a negative value indicating a deceleration. The coefficient of friction is used to adjust the effective force, confirming that the braking force is indeed opposing the rotation.
Kaxa2000
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A solid sphere w/ mass 100kg and radius 1m is spinning on axle. A brake pad is used to slow it down to a stop. While braking a force of 1N is applied on the pad. The coeff. of friction b/t pad and sphere is 0.5. The sphere is initially spinning at 100 rev/s, how long will it take to stop sphere?


Explanation on how to solve will be fine...
I'm guessing you find the moment of inertia of the sphere w/ the equation I = (2/5)MR^2 and then use constant acc techniques to find how long it takes to stop?
 
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Kaxa2000 said:
Explanation on how to solve will be fine...
I'm guessing you find the moment of inertia of the sphere w/ the equation I = (2/5)MR^2 and then use constant acc techniques to find how long it takes to stop?
Yes, if you actually mean constant angular acceleration techniques.
 
How do I get the third variable?
 
Which two variables do you have already?

You can figure out the angular acceleration from information in the problem statement. Two other variables are given to us directly.
 
Initial angular velocity = 100 rev/s
Final angular velocity = 0

I don't know how to get the angular acc.

Would you use the torque formula?

tau = Iα ?

I don't know what tau would be
 
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Anyone know?
 
Yes, use tau = I α

You can use the 1N force, and the coef. of friction, here.
 
I know tau = r x F

would you plug 1N for F? and 1 m(radius of sphere) for r?

tau = 1N x 1m = 1 N m

I = (2/5)(100kg)(1m)^2 = 40 kg m

so
angular acceleration = tau/I = (1/40) rad/s^2

is this right?
How would u use coeff of friction?
 
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The force that provides torque has to be in the direction opposing the rotation (or else how can the wheel stop?) The 1 N is applied perpendicular to the direction of rotation. How would you use the coefficient of friction to calculate the F in tau=r x F?
 
  • #10
Fk = ukN

so the Force would be (0.5)(1N)? = .5 N?
 
  • #11
Yeah, that's right.
 
  • #12
I'm getting a negative answer for time? Why is this

I used the

w - w0 = at

formula
 
  • #13
Does anyone know why I'm getting a negative t?

t = (w-w0)/(angular acc.)

(0 - 100)/(.0125 rad/s^2) = -8000 s
 
  • #14
The acceleration is negative in this case, since the sphere's rotation is slowing down and coming to a stop.
 
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