
A rigid massless rod of length $3l$has two masses attached at each end as shown in the figure. The rod is pivoted at point $P$ on the horizontal position, its instantaneous angular acceleration will be:
A) $\dfrac{g}{{2l}}$
B) $\dfrac{{7g}}{{3l}}$
C) $\dfrac{g}{{13l}}$
D) $\dfrac{g}{{3l}}$
Answer
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Hint:The rod is in the horizontal position and the instantaneous angular acceleration will be calculated by the torque formula. The magnitude of the torque will be produced by the force and also we need to apply the second law of motion to find the solution.
Formula used:
$\tau = rF\sin \theta $
Where,
$\tau $Is the torque
$r$Is the radius
$F$Is the force
$\theta $Is Angle between force and moment arm.
Complete step by step answer:
Torque can be measured by the force that causes an object to rotate about an axis in linear motion, the torque causes an object to acquire the angular acceleration.
$\tau = F.r\sin (\theta )$
Where,
$r$is the length of the moment arm.
$\theta $is the angle between the force and the moment arm.
If the force is at right angle to the moment of the arm then the sine becomes 1 as,
$\tau = r.F$
Where this is also calculated in either form,
$\tau = {r_1}.F$
$\tau = r.{F_1}$
Where,
${r_1}$ is the perpendicular distance from the origin to the line of force
${F_1}$ is the component of force perpendicular to the line of joining the force.
According to second law of motion for pure rotation,
$\vec \tau = I\alpha $
The torque equation can applied about two point,
That is the centre of the motion and the point in which has zero velocity or the acceleration.
$ \Rightarrow $$3{M_0}2L - 5{M_0}gL = Id$
$ \Rightarrow $$I = 2{M_0}{(2l)^2} + 5{M_0}{l^2}$
$ \Rightarrow $$I = 13{M_0}{l^2}d$
$ \Rightarrow $$d = \dfrac{{ - {M_0}gl}}{{13{M_0}{l^2}}}$
$ \Rightarrow $$d = \dfrac{{ - g}}{{13l}}$
$ \Rightarrow $$d = \dfrac{g}{{13l}}$ due to anticlockwise
So therefore option (C) is correct.
Note:Torque is the application of Force on the rotational motion. Torque plays an important role in determining the force in which direction is required to carry out the task. Some of the objects experience torque like see-saws, wrenches etc.
Formula used:
$\tau = rF\sin \theta $
Where,
$\tau $Is the torque
$r$Is the radius
$F$Is the force
$\theta $Is Angle between force and moment arm.
Complete step by step answer:
Torque can be measured by the force that causes an object to rotate about an axis in linear motion, the torque causes an object to acquire the angular acceleration.
$\tau = F.r\sin (\theta )$
Where,
$r$is the length of the moment arm.
$\theta $is the angle between the force and the moment arm.
If the force is at right angle to the moment of the arm then the sine becomes 1 as,
$\tau = r.F$
Where this is also calculated in either form,
$\tau = {r_1}.F$
$\tau = r.{F_1}$
Where,
${r_1}$ is the perpendicular distance from the origin to the line of force
${F_1}$ is the component of force perpendicular to the line of joining the force.
According to second law of motion for pure rotation,
$\vec \tau = I\alpha $
The torque equation can applied about two point,
That is the centre of the motion and the point in which has zero velocity or the acceleration.
$ \Rightarrow $$3{M_0}2L - 5{M_0}gL = Id$
$ \Rightarrow $$I = 2{M_0}{(2l)^2} + 5{M_0}{l^2}$
$ \Rightarrow $$I = 13{M_0}{l^2}d$
$ \Rightarrow $$d = \dfrac{{ - {M_0}gl}}{{13{M_0}{l^2}}}$
$ \Rightarrow $$d = \dfrac{{ - g}}{{13l}}$
$ \Rightarrow $$d = \dfrac{g}{{13l}}$ due to anticlockwise
So therefore option (C) is correct.
Note:Torque is the application of Force on the rotational motion. Torque plays an important role in determining the force in which direction is required to carry out the task. Some of the objects experience torque like see-saws, wrenches etc.
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