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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.

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onyebuchinnaji

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

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

(b) Applied torque is 0.071813 Nm.

Calculation of the moment of inertia and torque:

Since

The radius of the grinding wheel, R = 8.5cm

Mass of the grinding wheel, m = 0.580kg

Now

a.

Its moment of inertia about its center

I = ¹/₂MR²

I = ¹/₂(0.58)(0.085)²

I = 0.002095 kgm²

b.

Since

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

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

So,

Initial torque, τi = I x αi

Now

αi = ωi / t

αi = 157.1 / 5 = 31.42 rad/s²

And,

τi = 0.002095 x 31.42

τi = 0.06583 Nm

So,

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

So,

Applied torque = τi + τf

                        =  0.06583 Nm + 0.005983 Nm

                         = 0.071813 Nm

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