Question

A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored.If F3=0 and F1=12N, what does the magnitude of F2 have to be for there to be rotational equilibrium? If the L - shaped rod has a moment of inertia I=9 kg m2), F1=12N, F2=27, , and again F3=0, how long a time t would it take for the object to move through 45 degrees ( p/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Now consider the situation in which F1=12N and F2=0, but now a force with nonzero magnitude F3 is acting on the rod. What does F3 have to be to obtain equilibrium?

A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored.

If F3=0 and F1=12N, what does the magnitude of F2 have to be for there to be rotational equilibrium?

If the L-shaped rod has a moment of inertia I=9 kg m2), F1=12N, F2=27, , and again F3=0, how long a time t would it take for the object to move through 45 degrees ( p/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object.

Now consider the situation in which F1=12N and F2=0, but now a force with nonzero magnitude F3 is acting on the rod. What does F3 have to be to obtain equilibrium?

Answer 1

Concepts and reason

The concept is used to solve this equation in torque and momentum of inertia.

Torque:

In simple words, it can be defined as a rotational force which causes moment. Mathematically, it is defined as the product of force and the distance.

Moment of inertia:

It is a term representing an object's tendency to oppose angular acceleration.

Fundamentals

Static equilibrium condition:

$\begin{array}{l}\\\sum {{F_x} = 0} \\\\{\rm{The algebraic sum of the horizontal forces is zero}}{\rm{.}}\\\\\sum {{F_y}} = 0\\\\{\rm{The algebraic sum of the vertical forces is zero}}.\\\\\sum {{M_z}} = 0\\\\{\rm{The algebraic sum of the moments about a point is zero}}\\\end{array}$

Relation between torque $\tau$ and moment of inertia $a$ :

$\tau = \left( I \right)\left( \alpha \right)$

Here, $\alpha$ is the angular acceleration.

(1)

At the pivot, the torque of the system is equal to,

${\rm{External}}\;{\rm{torque,}}\tau = \left( {{F_1}} \right)\left( {0.03\;{\rm{m}}} \right)$

Substitute $0.36{\rm{ N}}$ for ${F_1}$ .

$\;\tau = 0.36\;{\rm{N}} \cdot {\rm{m}}\left( {{\rm{anti - clockwise}}} \right)$

Since the external torque is $0.36\;{\rm{N}} \cdot {\rm{m}}$ in anti-clockwise direction. Therefore torque of magnitude $0.36\;{\rm{N}} \cdot {\rm{m}}$ is applied in clockwise direction by means of force ${F_2}$ .

$\begin{array}{c}\\{F_2} = \frac{\tau }{{0.08\;{\rm{m}}}}\\\\ = \frac{{0.36\;{\rm{N}} \cdot {\rm{m}}}}{{0.08\;{\rm{m}}}}\\\\ = 4.5\;{\rm{N}}\\\end{array}$

(2)

At the pivot,

$\tau = \left( {{F_1}} \right)\left( {0.03\;{\rm{m}}} \right) - \left( {{F_2}} \right)\left( {0.08\;{\rm{m}}} \right)$

Substitute $12{\rm{ N}}$ for ${F_1}$ and $27{\rm{ N}}$ for ${F_2}$ .

$\tau = 1.8\;{\rm{N}} \cdot {\rm{m}}\left( {{\rm{clockwise}}} \right)$

From the expression of the torque,

$\tau = \left( I \right)\left( \alpha \right)$

Substitute $1.8\;{\rm{N}} \cdot {\rm{m}}$ for $\tau$ and $9{\rm{ kg}} \cdot {{\rm{m}}^2}$ for $I$ .

$\begin{array}{c}\\\alpha = \frac{{1.8\;{\rm{N}} \cdot {\rm{m}}}}{{9\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}}}\\\\ = 0.2\;\frac{{{\rm{rad}}}}{{{{\rm{s}}^{\rm{2}}}}}\\\end{array}$

Write the expression of the angular distance.

$\theta = ut + \frac{1}{2}\alpha {t^2}$ …… (1)

From the provided question the rod starts moving from rest therefore the initial speed of the object is $0{\rm{ m/s}}$ .

Substitute $\frac{\pi }{4}$ for $\theta$ and $0.2\;\frac{{{\rm{rad}}}}{{{{\rm{s}}^{\rm{2}}}}}$ for $\alpha$ in equation (1).

$\begin{array}{c}\\\frac{\pi }{4}\;{\rm{rad}} = \left( 0 \right)t + \frac{1}{2}\left( {0.2\;\frac{{{\rm{rad}}}}{{{{\rm{s}}^{\rm{2}}}}}} \right){t^2}\\\\{t^2} = \frac{{8\pi }}{{0.2}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

Take square root both sides.

$\begin{array}{c}\\t = \sqrt {7.85} \;{\rm{s}}\\\\ = 2.80\;{\rm{s}}\\\end{array}$

(3)

At the pivot, the expression for the torque is,

$\tau = \left( {{F_1}} \right)\left( {0.03\;{\rm{m}}} \right)$

Substitute $12{\rm{ N}}$ for ${F_1}$ .

$\begin{array}{c}\\\tau = \left( {12{\rm{ N}}} \right)\left( {0.03\;{\rm{m}}} \right)\\\\ = 0.36\;{\rm{N}} \cdot {\rm{m}}\left( {{\rm{anti - clockwise}}} \right)\\\end{array}$

Since the external torque is $0.36\;{\rm{N}} \cdot {\rm{m}}$ in anti-clockwise direction. Therefore torque of magnitude $0.36\;{\rm{N}} \cdot {\rm{m}}$ is applied in clockwise direction by means of force ${F_3}$ .

But ${F_3}$ has two component.

$\begin{array}{l}\\{\rm{Horizontal component}} = {F_3}\left( {\cos \;30^\circ } \right)\\\\{\rm{Vertical component}} = {F_3}\left( {\sin \;30^\circ } \right)\\\end{array}$

Since vertical component of ${F_3}$ contributes to the torque only. Therefore, force applied $\left( {{F_3}} \right)$ is expresses as follows,

$\begin{array}{c}\\{F_3}\left( {\sin \;30^\circ } \right) = \frac{\tau }{{0.08\;{\rm{m}}}}\\\\ = \frac{{0.36\;{\rm{N}} \cdot {\rm{m}}}}{{\left( {0.08\;{\rm{m}}} \right)\left( {\sin \;30^\circ } \right)}}\\\\ = 9\;{\rm{N}}\\\end{array}$

Ans: Part 1

The value of force ${F_2}$ to maintain rotational equilibrium is $4.5\;{\rm{N}}$

Part 2

The time taken by the rod to move through the angle $\frac{\pi }{4}$ is $2.8\;{\rm{s}}$ .

Part 3

The value of force ${F_3}$ to maintain rotational equilibrium is $9\;{\rm{N}}$