A machinist turns the power on to a grinding wheel, at rest, at time t = 0 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 38 rad/s. The wheel is run at that angular velocity for 30 s and then power is shut off. The wheel slows down uniformly at 2.1 rad/s2 until the wheel stops. In this situation, the angular acceleration of the wheel between t = 0 s and t = 10 s is

The angular acceleration of the grinding wheel at the given time interval is 3.8 rad/s^2

Explanation:

Given:

Initial time of the grinding wheel, t1 = 0

final time of the grinding wheel, t2 = 10 s

Initial angular velocity of the grinding wheel, u = 0

final angular velocity of the grinding wheel, v = 38 rad/s

The angular acceleration of the wheel between t = 0 s and t = 10 s, is calculated as;

a = dv/dt

Where;

a is angular acceleration

dv is change in angular velocity = v - u = 38 - 0 = 38 rad/s

dt is change in time = t2 - t1 = 10 -0 = 10

a = 38 / 10

a = 3.8 rad/s^2

Therefore, the angular acceleration of the grinding wheel at the given time interval is 3.8 rad/s^2

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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