
A particle travels in a circle of radius $20\;{\rm{cm}}$ at a speed that uniformly increases. If the speed increases from $5\;{\rm{m/s}}$ to $6\;{\rm{m/s}}$ in $2.0\;{\rm{s}}$, find the angular acceleration.
(A) $2.5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
(B) $3.5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
(C) $4.5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
(D) $5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
Answer
576.3k+ views
Hint: Here, we will use the formula for angular acceleration. The angular acceleration is the ratio of the change in the speed of a particle to the time in which the change in speed takes place. Since, the increase in the speed is uniform, then average tangential acceleration becomes equal to the instantaneous tangential acceleration. It is calculated by applying the formula for instantaneous acceleration.
Complete step by step answer:
Given: The radius of the circle is $r = 20\;{\rm{cm}}$, the initial speed of the particle is ${v_1} = 5\;{\rm{m/s}}$, the final speed of the particle is ${v_2} = 6\;{\rm{m/s}}$ and the time in which the speed of particle increases is $\Delta t = 2\;{\rm{s}}$
We know that the increment in the speed is uniform and due to this the average tangential acceleration is the same as the instantaneous tangential acceleration. Then, we write the formula to find the instantaneous tangential acceleration,
$a = \dfrac{{{v_2} - {v_1}}}{{\Delta t}}$
Now, we substitute all the given values in above relation to obtain the value of acceleration,
$
a = \dfrac{{6 - 5}}{2}\\
\implies a = \dfrac{1}{2}
$
Now, we perform the division of 1 by 2 to obtain the value of acceleration
$a = 0.5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
Therefore, the correct option from the above given options is option (A).
So, the correct answer is “Option A”.
Note:
The acceleration of a particle or object is a vector quantity. The acceleration of an object has both the direction and magnitude. There are various types of acceleration. Some of them are uniform, Non-uniform and instantaneous acceleration. We should not be confused between the types of acceleration. We should understand the concept and the type of acceleration given in the question and then apply the formula according to that.
Complete step by step answer:
Given: The radius of the circle is $r = 20\;{\rm{cm}}$, the initial speed of the particle is ${v_1} = 5\;{\rm{m/s}}$, the final speed of the particle is ${v_2} = 6\;{\rm{m/s}}$ and the time in which the speed of particle increases is $\Delta t = 2\;{\rm{s}}$
We know that the increment in the speed is uniform and due to this the average tangential acceleration is the same as the instantaneous tangential acceleration. Then, we write the formula to find the instantaneous tangential acceleration,
$a = \dfrac{{{v_2} - {v_1}}}{{\Delta t}}$
Now, we substitute all the given values in above relation to obtain the value of acceleration,
$
a = \dfrac{{6 - 5}}{2}\\
\implies a = \dfrac{1}{2}
$
Now, we perform the division of 1 by 2 to obtain the value of acceleration
$a = 0.5\;{\rm{rad/}}{{\rm{s}}^{\rm{2}}}$
Therefore, the correct option from the above given options is option (A).
So, the correct answer is “Option A”.
Note:
The acceleration of a particle or object is a vector quantity. The acceleration of an object has both the direction and magnitude. There are various types of acceleration. Some of them are uniform, Non-uniform and instantaneous acceleration. We should not be confused between the types of acceleration. We should understand the concept and the type of acceleration given in the question and then apply the formula according to that.
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