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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)