(A) The relationship between torque $(\tau)$,moment of inertia $(I)$,and angular acceleration $(\alpha)$ is given by the formula: $\tau = I \alpha$.Re

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(A) The relationship between torque $(\tau)$,moment of inertia $(I)$,and angular acceleration $(\alpha)$ is given by the formula: $\tau = I \alpha$.Rearranging for the moment of inertia,we get: $I = \frac{\tau}{\alpha}$.Given: $\tau = 31.4 \, N \cdot m$ and $\alpha = 4\pi \, rad/s^2$.Using $\pi \approx 3.14$,we have $\alpha = 4 \times 3.14 = 12.56 \, rad/s^2$.Substituting the values: $I = \frac{31.4}{12.56} = 2.5 \, kg \cdot m^2$.Thus,the moment of inertia is $2.5 \, kg \cdot m^2$.

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

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$A$ constant torque of $31.4 \, N \cdot m$ is applied to a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia of the wheel is ....... $kg \cdot m^2$. (in $.5$)

A
$2$
B
$3$
C
$4$
D
$5$

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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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$A$ constant torque of $31.4 \, N \cdot m$ is applied to a pivoted wheel. If the angular acceleration of the wheel is $4\pi \, rad/s^2$,then the moment of inertia of the wheel is ....... $kg \cdot m^2$. (in $.5$)

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

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