(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.Given $K = 750 \, J$ and $I = 2.4 \, kg \cdot m^2$,we have:$750 = \frac

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(B) The rotational kinetic energy $K$ is given by $K = \frac{1}{2} I \omega^2$.Given $K = 750 \, J$ and $I = 2.4 \, kg \cdot m^2$,we have:$750 = \frac{1}{2} \times 2.4 \times \omega^2$$750 = 1.2 \times \omega^2$$\omega^2 = \frac{750}{1.2} = 625$$\omega = \sqrt{625} = 25 \, rad/s$.Using the kinematic equation for rotational motion,$\omega = \omega_0 + \alpha t$,where $\omega_0 = 0$ (starting from rest):$25 = 0 + 5 \times t$$t = \frac{25}{5} = 5 \, s$.

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

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The moment of inertia of a body about a given axis is $2.4 \, kg \cdot m^2$. For what duration of time must an angular acceleration of $5 \, rad/s^2$ be applied about this axis to generate a rotational kinetic energy of $750 \, J$? (in seconds)

A
$6$
B
$5$
C
$4$
D
$3$

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System of Particles and Rotational Motion

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Relation between Torque and Angular acceleration and it's Application

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The moment of inertia of a body about a given axis is $2.4 \, kg \cdot m^2$. For what duration of time must an angular acceleration of $5 \, rad/s^2$ be applied about this axis to generate a rotational kinetic energy of $750 \, J$? (in seconds)

Similar Questions

$A$ disc of radius $0.4 \,m$ and mass $1 \,kg$ rotates about an axis passing through its center and perpendicular to its plane. The angular acceleration is $10 \,rad \,s^{-2}$. The tangential force applied to the rim of the disc is (in $\,N$)

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

$A$ 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 $I = \frac{24}{\pi} \, kg \cdot m^2$)

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