Question

slader A uniform cylindrical grinding wheel of mass 50.0 kg and diameter 1.0 m is turned on by an electric motor. The friction in the bearings is negligible. What torque must be applied to the wheel to bring it from rest to 120 rev/min in 20 revolutions?

Answer 1

Answer:

The value of torque on the rod is = 3.94 N - m

Explanation:

mass = 50 kg

Radius = 0.5 m

Moment of inertia ( I ) = $\frac{MR^{2} }{2}$

$I = 50 \frac{0.5^{2} }{2}$

I = 6.25 kg- $m^{2}$

Angular velocity $\omega$ = 120 $\frac{Rev}{min}$ = 12.6 $\frac{rad}{sec}$

Δ$\theta$ = No. of rev × 2$\pi$

Δ$\theta$ = 20 × 2$\pi$

Δ$\theta$ = 126 rad

From work energy theorem  

Work done is equal to change in kinetic energy.

$W_{AB} = K_{B} - K_{A}$ -------- (1)

Where A & B represents the initial & final states.

Since $K_{A}$ = 0

$W_{AB} = K_{B}$

$W_{AB} = \frac{1}{2} I \omega^{2}$

$W_{AB} = \frac{1}{2} 6.25 (12.6)^{2}$

$W_{AB} =$ 496 J

We know that

$W_{AB} =$ T Δ$\theta$

Where $W_{AB} =$ work done

T = torque applied on the rod

Δ$\theta$ = Angular displacement of the rod.

496 = T × 126

T = 3.94 N - m

Therefore the value of torque on the rod is = 3.94 N - m