मराठी

A disc of mass 10 kg and radius 0.1 m/s rotating at 120 r.p.m. A retarding torque brings it to rest in 10s.

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प्रश्न

A disc of mass 10 kg and radius 0.1 m/s rotating at 120 r.p.m. A retarding torque brings it to rest in 10s. If the same torque is due to force applied tangentially on the rim of the disc then magnitude of force is ______.

पर्याय

  • 0.1 πN

  • 0.2 πN

  • 0.4 πN

  • 0.8 πN

MCQ
रिकाम्या जागा भरा
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उत्तर

A disc of mass 10 kg and radius 0.1 m/s rotating at 120 r.p.m. A retarding torque brings it to rest in 10s. If the same torque is due to force applied tangentially on the rim of the disc then magnitude of force is 0.2 πN.

Explanation:

frequency f = 120 rpm `=120/60=2 "rps"`

ω = 2πf = 2π × 2 = 4π rad/s

Angular acceleration `=prop=(omega_2-omega_1)/t=(0-4pi)/10=(-2pi)/5  "rad/s"^2`

Moment of inertia `"I"="MR"^2/2=(10xx(0.1)^2)/2=0.05  "kg m"^2`

Torque `tau="I"alpha=0.05xx2/5pi=0.02pi  "Nm"`

τ = Fr

`therefore"F"=tau/r=(0.02pi)/0.1=0.2pi  "N"`

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Expression for Torque in Terms of Moment of Inertia
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