An object with a mass of 7 kg is traveling in a circular path of a radius of 9 m. If the object's angular velocity changes from 5 Hz to 6 Hz in 8 s, what torque was applied to the object?

2 Answers
Jane
Jan 13, 2018

442.26 Nm

Explanation:

Torque acting on a particle is given as, tau = Iprop
(Where, I is the moment if inertia and prop is angular acceleration)
Now, I = m(r^2)
Or, I= 567 SI unit
prop = (domega) / dt
= (omega final -omega initial)/t where omega is angular velocity
= (2pi/t) (n final - n initial) (as omega=2pin) where n is frequency
= 0.78 (rad)/(s^2)
Hence, tau = (567*0.78)N.m
Or,442.26 N.m

Narad T.
Jan 13, 2018

The torque is =445.3Nm

Explanation:

The torque is the rate of change of angular momentum

tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt

where I is the moment of inertia

The mass of the object is m=7kg

The radius of the path is r=9m

For the object, the moment of inertia is I=mr^2

So, I=7*(9)^2=567kgm^2

The rate of change of angular velocity is

(domega)/dt=(f_2-f_1)/t*2pi=(6-5)/8*2pi

=(1/4pi)rads^(-2)

So,

The torque is tau= 567*1/4pi=445.3Nm

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