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An object with a mass of # 3 kg# is traveling in a circular path of a radius of #2 m#. If the object's angular velocity changes from # 2 Hz# to # 17 Hz# in # 5 s#, what torque was applied to the object?

Answer 1

Leo Abeyta

The torque is #=226.2Nm#

The torque is the rate of change of angular momentum

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

The moment of inertia of the object is #I=mr^2#

#=3*2^2= 12 kgm^2#

The rate of change of angular velocity is

#(domega)/dt=(17-2)/5*2pi#

#=(6pi) rads^(-2)#

So the torque is #tau=12*(6pi) Nm=72piNm=226.2Nm#

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Answer 2

Julia Aguilar

To calculate torque (τ), use the formula:

$\tau = I \cdot \alpha$

where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. Moment of inertia for a point mass rotating about an axis at a distance $r$ is $I = m \cdot r^2$ .

Given:
Mass ( $m$ ) = 3 kg,
Radius ( $r$ ) = 2 m,
Initial angular velocity ( $\omega_i$ ) = 2 Hz,
Final angular velocity ( $\omega_f$ ) = 17 Hz,
Time ( $t$ ) = 5 s.

Angular acceleration ( $\alpha$ ) can be calculated using the formula:

$\alpha = \frac{\Delta \omega}{\Delta t}$

Substitute values into the formulas to find torque.

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Answer from HIX Tutor

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