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## 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 bearin

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?

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Physics
2 months
2021-08-05T10:24:29+00:00
2021-08-05T10:24:29+00:00 1 Answers
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## Answers ( )

Answer:The value of torque on the rod is = 3.94 N – mExplanation:mass = 50 kg

Radius = 0.5 m

Moment of inertia ( I ) =

I = 6.25 kg-Angular velocity= 120= 12.6Δ = No. of rev × 2

Δ = 20 × 2

Δ = 126 rad

From work energy theorem

Work done is equal to change in kinetic energy.

——– (1)Where A & B represents the initial & final states.

Since = 0

496 JWe know that

T ΔWhere work done

T = torque applied on the rod

Δ = Angular displacement of the rod.

496 = T × 126

T = 3.94 N – mTherefore the value of torque on the rod is = 3.94 N – m