(B) Initial angular velocity $\omega_1 = 720 \, rpm = \frac{720 \times 2\pi}{60} \, rad/s = 24\pi \, rad/s$.Final angular velocity $\omega_2 = 0 \, ra

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(B) Initial angular velocity $\omega_1 = 720 \, rpm = \frac{720 \times 2\pi}{60} \, rad/s = 24\pi \, rad/s$.Final angular velocity $\omega_2 = 0 \, rad/s$.Time taken $t = 8 \, s$.Using the equation of rotational motion $\omega_2 = \omega_1 - \alpha t$,where $\alpha$ is the angular retardation:$0 = 24\pi - \alpha(8) \implies \alpha = \frac{24\pi}{8} = 3\pi \, rad/s^2$.The magnitude of torque $\tau$ is given by $\tau = I\alpha$.Given $I = \frac{24}{\pi} \, kg \cdot m^2$,we have:$\tau = \left(\frac{24}{\pi}\right) \times (3\pi) = 72 \, N \cdot m$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

Medium

$A$ wheel is rotating about an axis through its centre at $720 \, rpm$. It is acted on by a constant torque opposing its motion for $8 \, s$ to bring it to rest finally. The value of torque (in $N \cdot m$) is (Given $I = \frac{24}{\pi} \, kg \cdot m^2$)

A
$48$
B
$72$
C
$96$
D
$120$

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Class 11 Questions

Class 11

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Chapter

System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

$A$ disc of mass $10 \ kg$ and radius $0.1 \ m$ is rotating at $120 \ rpm$. $A$ retarding torque brings it to rest in $10 \ s$. If the same torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is: (in $\pi \ N$)

MediumMHT CET 2020

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$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

Medium

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$A$ force of $1 \ N$ is applied at the free edge of a door of width $1.6 \ m$ to open it. If the door is to be opened by applying force at a point $0.4 \ m$ away from the hinges (axis of rotation),then the force required is ...... $N$.

Medium

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If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is:

MediumAIIMS 1997

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$A$ constant torque acting on a wheel changes its angular momentum from $A_0$ to $4A_0$ in $4 \ s$. The magnitude of the torque is:

Medium

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$A$ wheel is rotating about an axis through its centre at $720 \, rpm$. It is acted on by a constant torque opposing its motion for $8 \, s$ to bring it to rest finally. The value of torque (in $N \cdot m$) is (Given $I = \frac{24}{\pi} \, kg \cdot m^2$)

Similar Questions

$A$ disc of mass $10 \ kg$ and radius $0.1 \ m$ is rotating at $120 \ rpm$. $A$ retarding torque brings it to rest in $10 \ s$. If the same torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is: (in $\pi \ N$)

$A$ solid disc has a diameter of $0.5 \ m$ and a mass of $16 \ kg$. What torque must be applied to increase its angular velocity from $0$ to $120 \ rpm$ in $8 \ s$?

$A$ force of $1 \ N$ is applied at the free edge of a door of width $1.6 \ m$ to open it. If the door is to be opened by applying force at a point $0.4 \ m$ away from the hinges (axis of rotation),then the force required is ...... $N$.

If there is a change of angular momentum from $1\,J\cdot s$ to $5\,J\cdot s$ in $5\,s$,then the torque is:

$A$ constant torque acting on a wheel changes its angular momentum from $A_0$ to $4A_0$ in $4 \ s$. The magnitude of the torque is:

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