Would someone check my thinking please.
A solid sphere (mass, r, and moment of inertia given) is rotating at a given angular speed. Friction is constantly applied at the 'equator' of the sphere, and brings it to rest in 3s.
Find the force. I worked out the torque, and then rearranged: T=rxF to give the force. Will i be right if the maths is ok?
Then, if the same force is applied (same conditions) half way between equator and pole, will it take longer or shorter to stop?
Thanks.
yes on the first.
On the second, the torque needed is the same, however, it will need a larger force to make that same torque.
Thanks for that.
So it would take longer to stop then?
Yes, your approach is correct. To find the force applied, you need to calculate the torque first and then rearrange the torque formula to solve for the force. The torque (T) acting on the sphere can be calculated as the product of the moment of inertia (I) and the angular acceleration (α).
T = I * α
In this case, the sphere is being brought to rest, so the angular acceleration can be found by dividing the final angular speed (ω_f = 0) by the time taken (t = 3s).
α = ω_f / t = 0 / 3 = 0 rad/s^2
Since the sphere is rotating about its equator, the torque is constant throughout the process. Therefore, the torque (T) can also be written as the product of the applied force (F) and the radius of the sphere (r).
T = r * F
Now, equating the two torque equations, we get:
I * α = r * F
Simplifying the equation, we can solve for the force F:
F = (I * α) / r
So, your approach of rearranging the torque formula to solve for the force is correct.
Regarding the second part of your question, if the same force is applied halfway between the equator and the pole, it will take longer to bring the sphere to rest. This is because the sphere has a larger moment of inertia about the axis passing through the equator compared to the axis passing through the pole. Therefore, the angular acceleration will be smaller, resulting in a longer time to stop the rotation.
I hope this explanation helps! Let me know if there's anything else I can assist you with.