A grinding wheel is a uniform cylinder with a radius of 8.5cm and a mass of 0.580kg. Calculate

a. its moment of inertia about its center.
b. the applied torque needed to accelerate it from rest to 1500rpm in 5.00s if it is known to slow down from 1500rpm to rest in 55.0s.

(a) its moment of inertia about its center is 0.002095 kgm²

(b) Applied torque is 0.071813 Nm

Explanation:

Given;

Radius of the grinding wheel, R = 8.5cm

Mass of the grinding wheel, m = 0.580kg

Part (a) its moment of inertia about its center

I = ¹/₂MR²

I = ¹/₂(0.58)(0.085)²

I = 0.002095 kgm²

Part (b)

Given;

initial angular velocity, ωi = 1500rpm = 157.1 rad/s

final angular velocity, ωf = 1500rpm = 157.1 rad/s

Initial torque, τi = I x αi

αi = ωi / t

αi = 157.1 / 5 = 31.42 rad/s²

τi = 0.002095 x 31.42

τi = 0.06583 Nm

Final torque, τf =  I x αf

αf = ωf / t

αf = 157.1 / 55 = 2.856 rad/s²

τf = 0.002095 x 2.856

τf = 0.005983 Nm

Applied torque = τi + τf

                          =  0.06583 Nm + 0.005983 Nm

                          = 0.071813 Nm

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

b would be the answer