
A constant torque of $ 31.4{\text{ N - m}} $ is exerted on a pivoted wheel .If the angular acceleration of the wheel is $ 4\pi {\text{ }}\dfrac{{{\text{rad}}}}{{{s^2}}} $ ,then the moment of inertia of the wheel is:
$ {\text{A}}{\text{. }}2.5{\text{ kg - }}{{\text{m}}^2} \\ $
$ {\text{B}}{\text{. }}3.5{\text{ kg - }}{{\text{m}}^2} \\ $
$ {\text{C}}{\text{. }}4.5{\text{ kg - }}{{\text{m}}^2} \\ $
$ {\text{D}}{\text{. }}5.5{\text{ kg - }}{{\text{m}}^2} \\ $
Answer
484.5k+ views
Hint: We can use the relation between torque, momentum and moment of inertia to solve the problem. Substitute the value of pi and solve it using the relation .
$ \tau = I\alpha $
Complete step by step answer:
Torque, Moment of Inertia and Angular Acceleration
Torque is the rotational equivalent of Force. Similarly, Moment of Inertia is the rotational equivalent of Mass and Angular Acceleration is the rotational equivalent of Linear Acceleration.
In rotational motion, torque is required to produce an angular acceleration of an object. The amount of torque required to produce an angular acceleration depends on the distribution of the mass of the object. The moment of inertia is a value that describes the distribution. It can be found by integrating over the mass of all parts of the object and their distances to the center of rotation, but it is also possible to look up the moments of inertia for common shapes. The torque on a given axis is the product of the moment of inertia and the angular acceleration. The units of torque are Newton-meters (N∙m).
Just like F = ma, in Rotational Mechanics we have:
$ \tau = I \times \alpha $
Where:
$ \tau $ = Torque
$ I $ = Moment of Inertia
$ \alpha $ = Angular Acceleration
Here, we have our data:
$ \tau = 31.4 Nm $
$ \alpha = 4\pi rad/s^2 $
We need the Moment of Inertia (I).
$ \Rightarrow I = \dfrac{\tau }{\alpha } = \dfrac{{31.4}}{{4 \times 3.14}} = 2.5{\text{ kg - }}{{\text{m}}^2} $ as $ \pi = 3.14 $
Therefore, the correct answer is option A.
Note:
Torque is equal to the moment of inertia times the angular acceleration. The larger the moment of inertia, the smaller the angular acceleration. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates.
$ \tau = I\alpha $
Complete step by step answer:
Torque, Moment of Inertia and Angular Acceleration
Torque is the rotational equivalent of Force. Similarly, Moment of Inertia is the rotational equivalent of Mass and Angular Acceleration is the rotational equivalent of Linear Acceleration.
In rotational motion, torque is required to produce an angular acceleration of an object. The amount of torque required to produce an angular acceleration depends on the distribution of the mass of the object. The moment of inertia is a value that describes the distribution. It can be found by integrating over the mass of all parts of the object and their distances to the center of rotation, but it is also possible to look up the moments of inertia for common shapes. The torque on a given axis is the product of the moment of inertia and the angular acceleration. The units of torque are Newton-meters (N∙m).
Just like F = ma, in Rotational Mechanics we have:
$ \tau = I \times \alpha $
Where:
$ \tau $ = Torque
$ I $ = Moment of Inertia
$ \alpha $ = Angular Acceleration
Here, we have our data:
$ \tau = 31.4 Nm $
$ \alpha = 4\pi rad/s^2 $
We need the Moment of Inertia (I).
$ \Rightarrow I = \dfrac{\tau }{\alpha } = \dfrac{{31.4}}{{4 \times 3.14}} = 2.5{\text{ kg - }}{{\text{m}}^2} $ as $ \pi = 3.14 $
Therefore, the correct answer is option A.
Note:
Torque is equal to the moment of inertia times the angular acceleration. The larger the moment of inertia, the smaller the angular acceleration. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates.
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