(A) The angular position is given by $\theta(t) = 5t^2 - 8t$.Angular velocity $\omega$ is the derivative of angular position with respect to time: $\o

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(A) The angular position is given by $\theta(t) = 5t^2 - 8t$.Angular velocity $\omega$ is the derivative of angular position with respect to time: $\omega = \frac{d\theta}{dt} = 10t - 8$.Angular acceleration $\alpha$ is the derivative of angular velocity with respect to time: $\alpha = \frac{d\omega}{dt} = 10 \text{ rad/s}^2$.The moment of inertia of a circular disc about an axis perpendicular to its plane and passing through its center is $I = \frac{1}{2}MR^2$.The torque applied is $\tau = I\alpha = (\frac{1}{2}MR^2)(10) = 5MR^2$.Power $P$ delivered by the torque is $P = \tau \omega$.At $t = 2$ s,$\omega = 10(2) - 8 = 12 \text{ rad/s}$.Therefore,$P = (5MR^2)(12) = 60 MR^2$ $W$.

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

Medium

$A$ circular disc of radius $R$ meter and mass $M$ kg is rotating around the axis perpendicular to the disc. An external torque is applied to the disc such that $\theta(t) = 5t^2 - 8t$,where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque,when $t = 2$ s (in $MR^2$)?

JEE MAIN 2025 All JEE MAIN

A
$60$
B
$72$
C
$108$
D
$8$

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System of Particles and Rotational Motion

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Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \, J$,an angular acceleration of $25 \, rad/s^2$ must be applied about that axis for a duration of ........ $s$.

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$A$ thin uniform circular disc of mass $\frac{10}{\pi^2} \,kg$ and radius $2 \,m$ is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from $90 \,rev/min$ to $120 \,rev/min$ is (in $\,J$)

MediumTS EAMCET 2025

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The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

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$A$ thin,uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of $(g = \text{acceleration due to gravity})$

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Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

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Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

Similar Questions

The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \, J$,an angular acceleration of $25 \, rad/s^2$ must be applied about that axis for a duration of ........ $s$.

$A$ thin uniform circular disc of mass $\frac{10}{\pi^2} \,kg$ and radius $2 \,m$ is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from $90 \,rev/min$ to $120 \,rev/min$ is (in $\,J$)

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

$A$ thin,uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of $(g = \text{acceleration due to gravity})$

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$. Column-$I$ Column-$II$ $(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$ $(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$ $(c)$ $L = I \omega$

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Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$