$A$ wheel having a moment of inertia $2 \; kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \; rpm$ about this axis. The torque required to stop the wheel's rotation in one minute is:

  • A
    $ \frac{2\pi}{15} \; N \cdot m $
  • B
    $ \frac{\pi}{12} \; N \cdot m $
  • C
    $ \frac{\pi}{15} \; N \cdot m $
  • D
    $ \frac{\pi}{18} \; N \cdot m $

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Similar Questions

Which of the following are correct expressions for torque acting on a body?
$A. \ \vec{\tau}=\vec{ r } \times \vec{ L }$
$B. \ \vec{\tau}=\frac{ d }{ dt }(\vec{ r } \times \vec{ p })$
$C. \ \vec{\tau}=\vec{ r } \times \frac{ d \vec{ p }}{ dt }$
$D. \ \vec{\tau}= I \vec{\alpha}$
$E. \ \vec{\tau}=\vec{ r } \times \vec{ F }$
($\vec{ r }=$ position vector; $\vec{ p }=$ linear momentum;
$\vec{ L }=$ angular momentum; $\vec{\alpha}=$ angular acceleration;
$I=$ moment of inertia; $\vec{ F }=$ force; $t =$ time)
Choose the correct answer from the options given below:

The moment of inertia of a wheel rotating about an axis is $3 \times 10^2 \ kg \ m^2$ and it has a constant angular speed of $4.6 \ rad \ s^{-1}$. If a retarding torque of $6.9 \times 10^2 \ N \ m$ is applied to the wheel,in how many seconds will the wheel come to rest?

$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is:

What constant force must be applied tangentially at the equator to stop the Earth's rotation in one day?

In the given figure,$r_1$ and $r_2$ are $10 \ cm$ and $20 \ cm$ respectively. If the moment of inertia of the wheel is $1500 \ kg \cdot m^2$,find the angular acceleration.

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