A wheel of radius R is free to rotate about its own axis. A tangential force F is applied on the wheel along its rim. If q is the angular displacement of wheel due to the force F, then work done by F is:
A) $RFq$
B) $\dfrac{{RF}}{q}$
C) $\dfrac{{qF}}{R}$
D) $\dfrac{{qR}}{F}$
Answer
654.3k+ views
Hint:A wheel of radius R is rotating and hence there will be some torque applied on the system. It is the cross product of the tangential force applied to the radius of the wheel (in this case). If torque is applied then there must be a work done to rotate the rim which is equal to the sum of all the torques integrated with angular displacement.
Complete solution:
Torque is a measure of how much force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'.
Mathematically, torque can be written as $\tau $ = F × r ×$\sin \theta $
It has units of Newton-meters. When the sum of all torques acting on an object equals zero, it is in rotational equilibrium.
We are given:
A wheel of radius R is free to rotate about its own axis, and a tangential force F is applied on the wheel along its rim.
Here, q is the angular displacement of wheel due to force F
We need to find the work done by the force.
Force is applied at the rim then torque is written as the cross product of the tangential force and radius.
$ \rightarrow \tau = F \times R$. The cross product a × b is defined as a vector c that is perpendicular to both F and R, with a direction given by the right-hand rule and equal magnitude.
Now solving the cross product, $\tau = FR\sin \theta $ here, $\theta $ is the displacement ………. (1)
The total work done to rotate a rigid body through an angle $\theta $ about a fixed axis is the sum of the torques integrated over the angular displacement.
We are given that the angular displacement $d\theta $ is q
Then the work done by F is $\tau d\theta $
Substituting the value of torque and angular displacement we get, W=RFq
Hence, option A is the correct answer.
Note:Angular displacement of a body is the angle in radians (degrees, revolutions) through which a point revolves around a centre or line has been rotated in a specified sense about a specified axis. It is only applicable in rotational mechanics. For linear mechanics linear displacement is used. It is the movement in one direction along a single axis.
Complete solution:
Torque is a measure of how much force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'.
Mathematically, torque can be written as $\tau $ = F × r ×$\sin \theta $
It has units of Newton-meters. When the sum of all torques acting on an object equals zero, it is in rotational equilibrium.
We are given:
A wheel of radius R is free to rotate about its own axis, and a tangential force F is applied on the wheel along its rim.
Here, q is the angular displacement of wheel due to force F
We need to find the work done by the force.
Force is applied at the rim then torque is written as the cross product of the tangential force and radius.
$ \rightarrow \tau = F \times R$. The cross product a × b is defined as a vector c that is perpendicular to both F and R, with a direction given by the right-hand rule and equal magnitude.
Now solving the cross product, $\tau = FR\sin \theta $ here, $\theta $ is the displacement ………. (1)
The total work done to rotate a rigid body through an angle $\theta $ about a fixed axis is the sum of the torques integrated over the angular displacement.
We are given that the angular displacement $d\theta $ is q
Then the work done by F is $\tau d\theta $
Substituting the value of torque and angular displacement we get, W=RFq
Hence, option A is the correct answer.
Note:Angular displacement of a body is the angle in radians (degrees, revolutions) through which a point revolves around a centre or line has been rotated in a specified sense about a specified axis. It is only applicable in rotational mechanics. For linear mechanics linear displacement is used. It is the movement in one direction along a single axis.
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