Answer:
(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