Grindstone: rotational dynamics and kinematics you have a grindstone (a disk) that is 133.0 kg, has a 0.635 m radius, and is turning at 95 rpm, and you press a steel axe against it with a radial force of 32 n. (a) assuming the coefficient of kinetic friction between the steel and stone is 0.10. calculate the magnitude of the kinetic frictional force (in newtons) on the grindstone. hint: the applied radial force is a normal force on the grindstone.

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where N can be obtained as follows:

∑ Fc = m*ac   ⇒   N - F = m*ac = m*ω²*R    ⇒  N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

the cellphone, they would then know if the killer would have contacted them or always hated them. or if it was a breakup. but if that is not the case, then it would be the clothing. to put in a bullet, it would have to be put in by hands. hands have finger prints which, would have giving you the answer of who done it.

explanation:

unbalanced force

explanation: