Question

A particle of mass $10g$ moves along a circle of radius $6.4cm$ with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $8\times {{10}^{-4}}J$ by the end of the second revolution after the beginning of the motion?
A.) $0.1m/{{s}^{2}}$
B.) $0.15m/{{s}^{2}}$
C.) $0.18m/{{s}^{2}}$
D.) $0.2m/{{s}^{2}}$

Answer 1

Hint: This problem can be solved by finding out the final velocity at the end of the second rotation from the kinetic energy given and then use the equations of uniform motion to find the acceleration using the change in velocity and the distance travelled by the body. The distance travelled by the body can be calculated as the product of the number of rotations and the perimeter of the circle for each rotation.
Formula used:

$KE=\dfrac{1}{2}m{{v}^{2}}$
where $KE$ is the kinetic energy of a body of mass $m$ moving at a velocity $v$.

Distance covered by a body while going around a circle $\left( s \right)$ is given by
$\text{No}\text{. of rotations }\left( N \right)\text{ made by the body around the circle}\times \text{circumference of the circle}$
$\text{Circumference of the circle = }2\pi R$
where $R$ is the radius of the circle.

$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$ (Equation of uniform motion)
Where $v$ and $u$ are the final and initial velocities of the body, $s$ is the displacement during the velocity change and $a$ is the acceleration which caused the velocity change.

Complete step by step answer:
We will solve this problem by using the concept that the kinetic energy of a body depends upon its translational velocity. We will find out the velocity of the body from the given kinetic energy and then use the equations of uniform motion to find out the acceleration of the body using this velocity and the total displacement covered by the body while going around the circle.
The kinetic energy of a body is given by
$KE=\dfrac{1}{2}m{{v}^{2}}$ --(1)
where $KE$ is the kinetic energy of a body of mass $m$ moving at a velocity $v$.

Distance covered by a body while going around a circle $\left( s \right)$ is given by
$\text{No}\text{. of rotations }\left( N \right)\text{ made by the body around the circle}\times \text{circumference of the circle}$ -(2)

$\text{Circumference of the circle = }2\pi R$ --(3)
where $R$ is the radius of the circle.

$\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}=a$ (Equation of uniform motion) --(4)
Where $v$ and $u$ are the final and initial velocities of the body, $s$ is the displacement during the velocity change and $a$ is the acceleration which caused the velocity change.

Therefore, let us analyse the question.

Given, final kinetic energy of the body $\left( {{K}_{f}} \right)=8\times {{10}^{-4}}J$
Let the final velocity of the body be $v$.

The given mass of the body is $m=10g={{10}^{-2}}Kg$ $\left( \because 1g={{10}^{-3}}Kg \right)$
Hence, plugging in these values in (1), we get,

$8\times {{10}^{-4}}=\dfrac{1}{2}\times {{10}^{-2}}\times {{v}^{2}}$
$\therefore {{v}^{2}}=16\times {{10}^{-2}}$
Now, doing square root on both sides, we get,
$\sqrt{{{v}^{2}}}=\sqrt{16\times {{10}^{-2}}}$
$\therefore v=4\times {{10}^{-1}}=0.4m/s$ --(5)

Now, the initial velocity of the body at the beginning of the motion is considered to be zero.

Hence, initial velocity $u=0$. --(6)

Now, radius of the circular path is $R=6.4cm=6.4\times {{10}^{-2}}m$ $\left( \because 1cm={{10}^{-2}}m \right)$
Therefore, using (3), perimeter of the circular path is $2\pi \times 6.4\times {{10}^{-2}}=40.21\times {{10}^{-2}}m$ --(7)

The total number of rotations covered by the body according to the question is $N=2$--(8).

Now, plugging in (7) and (8), we get the total distance travelled by the body as

$s=2\times 40.21\times {{10}^{-2}}m=80.42\times {{10}^{-2}}m$ --(9)

Now, to get the acceleration of the body, we will use equation (4).
Therefore, plugging in (5), (6) and (9) in (4), we get,

$\dfrac{{{\left( 4\times {{10}^{-1}} \right)}^{2}}-{{0}^{2}}}{2\times 80.42\times {{10}^{-2}}}=a$
$\therefore \dfrac{16\times {{10}^{-2}}}{160.84\times {{10}^{-2}}}=a$
$\therefore a\approx 0.1m/{{s}^{2}}$
Hence, the required acceleration is $0.1m/{{s}^{2}}$.
Therefore, the correct option is A) $0.1m/{{s}^{2}}$.

Note: Students might get confused upon reading the term tangential acceleration. However, tangential acceleration is the linear acceleration that tends to change the linear speed of the body. However, if the question stated that the angular acceleration was angular acceleration then, the question would have changed completely. Then the angular acceleration would have to be converted into its corresponding tangential acceleration or else, the velocity would have to be changed to the angular velocity using the value of the radius of the path.Here, we have solved the problem by considering all the variables for that of linear motion since the final answer requires one that pertains to linear motion (tangential acceleration). We could also have solved this for rotational motion by converting all the quantities into their rotational analogues (mass to moment of inertia, velocity to angular velocity and displacement to angular displacement). Then the final answer that we would have got would be of angular acceleration which we would then have to convert back to tangential acceleration.