Home
Class 11
PHYSICS
A balance-wheel of a watch has a moment ...

A balance-wheel of a watch has a moment of inertia equal to `2 xx 10^(-8)` kg `m^(2)`. When it is rotated by `30^(@)`, a restoring torque of 5.13 `xx 10^(-6)` Nm is generated, calculate its frequency of oscillations.

Text Solution

Verified by Experts

Given, `theta =30^(@) = pi//6` rad
`tau = 5.13 xx 10^(-6)` Nm
`I = 2 xx 10^(-8)kg m^(2)`
Restoring torque per unit angular displacement (torsional constant) is
`C = (tau)/(theta) = (5.13xx10^(-6))/(pi//6) = (5.13 xx 10^(-6) xx 6 xx 7)/(22)`
` = 9.8 xx 10^(-6)` Nm `rad^(-1)`
`therefore` Frequency of oscillations
`v = (1)/(2pi)sqrt((C)/(I)) = (1)/(2xx3.14) sqrt((9.8 xx 10^(-6))/(2xx10^(-8)))`
` = (1)/(6.28) sqrt(4.9 xx 10^(2)) = ( 7 xx sqrt(10))/(6.28)`
= 3.52 Hz
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    MODERN PUBLICATION|Exercise PRACTICE PROBLEMS|15 Videos

  • OSCILLATIONS

    MODERN PUBLICATION|Exercise PRACTICE PROBLEMS 2|20 Videos

  • MOTION IN A STRAIGHT LINE

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|16 Videos

  • PHYSICAL WORLD

    MODERN PUBLICATION|Exercise Revision exercises (Long answer questions)|6 Videos

Similar Questions

Explore conceptually related problems

A wheel of mass 8 kg has moment of inertia equals to 0.5 "kg-m"^(2) . Determine its radius of gyration.

The balance wheel of a watch has a moment of inertia of 2xx10^(-8)kgm^(2) and the torsional constant of its hair spring is 9.8xx10^(-8) Nm rad^(-1) calculate its frequency.

A wheel has moment of inertia 5 xx 10^(-3) kg m^(2) and is making 20 "rev" s^(-1) . The torque needed to stop it in 10 s is….. xx 10^(-2) N-m

A wheel of mass 10 kg has a moment of inertia of 160 kg-m^(2) about its own axis, the radius of gyration will be

A wheel has a moment of inertia of 5xx10^(-3) kg m"^(2) and is making "20 rev/s". What is the magnitude of the torque required to stop it in 10 s?

A flywheel of moment of inertia 3 xx 10^(2) kg m^(2) is rotating with uniform angular speed of 4.6 rad s^(-1) . If a torque of 6.9 xx 10^(2)Nm retards the wheel, then the time in which the wheel comes to rest is

A magnetic dipole of magnetic moment 6xx 10^(-2) A m^2 and moment of inertia 12 xx 10^(-6) kg m^2 performs oscillations in a magnetic field of 2 xx 10^(-2) T. The time taken by the dipole to complete 20 oscillations is (pi~~3)

The moment of inertia of a flywheel is 4 kg m^(2) . What angular acceleration will be produced in it by applying a torque of 10 Nm on it?

The torque required to stop a wheel having moment of inertia 5xx10^(-3)kgm^(2) from a speed of 10 radian s^(-1) in 20s is