1.159 Work done in uniform circular motion

1.159 Work done in uniform circular motion

Problem 1.159:.A body of mass $‘m’$ is moving in a circle of radius with a constant speed $‘v’$. Calculate the work done by the resultant force in moving the body over half circumference and full circumference respectively.

Solution:

This is motion along a circular path. Resultant force on the particle produces the necessary centripetal force.

problem-1-159

Centripetal force is directed toward the center of the path and velocity is tangential to the path.

Therefore  resultant force is always perpendicular to the velocity.

This makes work done zero as there is no change in speed and kinetic energy of the body.

Then the work done by the resultant force in moving the body over half circumference and full circumference respectively are zero.

1.158 Angular acceleration of a disc

1.158 Angular acceleration of a disc

Problem 1.158: The disk has a mass $M$ and a radius $R$. If a block of mass $m$ is attached to the cord, determine the angular acceleration of the disk when the block is released from rest. 

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Solution:

When released, the block falls down vertically due to its weight $(mg)$. This develops a tension$(T)$ in the string which acts upward on the block.

The tension in the cord produces a torque on the disc.

The cord unwinds and this makes the disc to rotate about its own axis.

While falling downward he block losses its gravitational potential energy and acquires kinetic energy due to its linear motion.

The disc acquires rotational kinetic energy due to its rotational motion.3

Writing the force equation for the block we get, $mg\:-\:T\:=\:ma$ where $a$ is the linear acceleration of the block.

Writing the torque equation for the disc we get, $TR\:=\:I\alpha$. Here $I$ is the moment of inertia of the disc about its axis of rotation and $\alpha$ is the angular acceleration.

Moment of inertia of the disc about its axis of rotation is $\frac{MR^2}{2}$.

Angular acceleration of the disc.

When the cord unwinds from the disc, disc rotates.

If the cord does not slip on the disc, then angular acceleration is related to linear acceleration of the block as $\alpha \:=\:\frac{a}{R} $.

Therefore, $TR\:=\: \frac{MR^2}{2}\times\frac{a}{R} $.

Or, $T\:=\: \frac{Ma}{2} $.

Using this value of tension in the first equation $mg\:-\:T\:=\:ma$  we can get, $mg\:-\:\frac{Ma}{2}\:=\:ma$

Or, $mg\:=\:\frac{Ma}{2}\:+\:ma\:=\:a(\frac{M}{2}\:+\:m)\:=\:\frac{M+2m}{2}\times a$.

Solving we get linear acceleration of the block as $a\:=\:\frac{2mg}{M+2m}$.

Angular acceleration of the disc, $\alpha\:=\:\frac{a}{R}\:=\:\frac{2mg}{(M+2m)R}$.

1.157 Force on a rod from a wall

1.157 Force on a rod from a wall

Problem 1.157: A rod of mass $M$ is hinged at one end and is balanced horizontally with the help of a string as shown in the diagram. Determine the force exerted by the wall on the rod.

 18-Nov-2014 Problem 1.156

Solution:

The different forces acting on the rod are: weight $Mg$, tension $T$ in the string and normal reaction $N$ from the wall.

18-Nov-2014 Problem 1.156 B

The tension is resolved into vertical and horizontal components as shown below.

 

 

18-Nov-2014 Problem 1.156 C

For vertical equilibrium, $T\:sin\:30\:=\:Mg$.

Therefore, $T\:=\:\frac{Mg}{sin\:30}\:=\:\frac{Mg}{0.5}\:=\:2Mg$.

So the tension in the string is two times the weight of the rod.

For horizontal equilibrium, $T\:cos\:30\:=\:N$.

Therefore the normal reaction, $N\:=\:2Mg\:\frac{\sqrt{3}}{2}\:=\:\sqrt{3}mg$.

The wall exerts a force which is $\sqrt{3}$ times the weight of the rod.

1.156 Tension in a string

1.156 Tension in a string

Problem 1.156: A rod of mass $M$ is hinged at one end and is balanced horizontally with the help of a string as shown in the diagram. Determine the tension in the string.

 18-Nov-2014 Problem 1.156 B

Solution:

The different forces acting on the rod are: weight $Mg$, tensi

on $T$ in the string and normal reaction $N$ from the wall.

18-Nov-2014 Problem 1.156

The tension is resolved into vertical and horizontal components as shown below.

18-Nov-2014 Problem 1.156 C 

For vertical equilibrium, $T\:sin\:30\:=\:Mg$.

Therefore, $T\:=\:\frac{Mg}{sin\:30}\:=\:\frac{Mg}{0.5}\:=\:2Mg$.

So the tension in the string is two times the weight of the rod.

1.155 Work done and change in energy

1.155 Work done and change in energy

Problem 1.155: A thin rod of mass $M$ and length $L$ is lying on a horizontal surface with one end hinged to the ground. Determine the work done to lift the other end through a distance $\frac{L}{2}$. 

17-Nov-2014 Problem 1.155 A

Solution:

Here work done is equal to the change in gravitational potential energy of the center of mass of the system.

Center of mass of the rod is at its midpoint.

For the thin rod in the problem, center of mass is initially at the ground level.

When one end is raised without disturbing the other end, the center of mass also rises. This is shown in the following diagram. 

17-Nov-2014 Problem 1.155 B

Let us find the height through which the center of mass is raised.

This is marked as $h$ in the following diagram. 

17-Nov-2014 Problem 1.155 C

In the following diagram, we take two triangles namely OAD and OBC.

17-Nov-2014 Problem 1.155 D

From triangle OAD, we get  $sin\:\theta\:=\:\frac{h}{\frac{L}{2}}$ and from triangle OBC we get $sin\:\theta\:=\:\frac{\frac{L}{2}}{L}$.

Or, $\frac{h}{\frac{L}{2}}\:=\:\frac{\frac{L}{2}}{L}$.

Or, $\frac{2h}{L}\:=\:\frac{1}{2}$.

Or, $h\:=\:\frac{L}{4}$.

So the center of mass rises through $\frac{L}{4}$.

Change in the potential energy of the rod is $Mgh\:=\:Mg\frac{L}{4}\:=\:\frac{MgL}{4}$.

This is equal to the work done in raising the rod.

1.154 Moment of inertia

1.154 Moment of inertia

Problem 1.154: Four identical discs of mass $M$ and radius $R$ are arranged as shown in the diagram. Determine the moment of inertia of the system about the axis $X-X’$.

16-Nov-2014 Problem 1.154 A

Solution:

Moment of inertia of the system is the sum of moment of inertia of each disc about the same axis.

Let us consider the first disc.

16-Nov-2014 Problem 1.154 B

Moment of inertia of the disc about the axis $X-X’$ can be determined using the parallel axis theorem. For this another axis parallel to the given axis and passing through the center of mass must b considered. This axis is $Y-Y’$.

16-Nov-2014 Problem 1.154 C

Now if $I_x$ is moment of inertia about the given axis and  $I_y$ is the moment of inertia about the parallel axis then, $I_x\:=\:I_y\:+\:Md^2$ where $M$ is the mass and $d$ is the distance between the two axes.

The axis $Y-Y’$  is along the diameter of the disc. So $I_y\:=\:\frac{MR^2}{2}$.

Here distance between the two axes $d$  is equal to the radius $R$ of the disc.

So, $I_x\:=\:\frac{MR^2}{2}\:+\:MR^2\:=\:\frac{3}{2}MR^2$.

So the moment of inertia of one disc about the given axis is $\frac{3}{2}MR^2$.

All the other discs being identical will also have the same moment of inertia.

Therefore the total moment of inertia is $4\times\frac{3}{2}MR^2\:=\:6MR^2$.