Question

If a cycle wheel of radius 4m completes one revolution in 2 seconds. Then the acceleration of the cycle is,
A) ${\text{4}}{{\pi }^{\text{2}}}{\text{m/}}{{\text{s}}^{\text{2}}}$.
B) ${\text{2}}{{\pi }^{\text{2}}}{\text{m/}}{{\text{s}}^{\text{2}}}$.
C) ${{\pi }^{\text{2}}}{\text{m/}}{{\text{s}}^{\text{2}}}$.
D) ${\text{4m/}}{{\text{s}}^{\text{2}}}$

Answer 1

Hint
Use the formula ${a_c} = {\omega ^2}R$ to calculate the acceleration and also replace the value of angular frequency in terms of Time period by using the formula $\omega = \dfrac{{2\pi }}{T}$ and solve for acceleration to get the answer.

Complete step by step answer
It is given that the time period of the cycle wheel = 2 second.
Also given that the radius of the cycle wheel is = 4 m.
We know that the centripetal acceleration of the cycle is given by the formula,
${a_c} = {\omega ^2}R$ where,
$\omega $ is the angular frequency of the cycle wheel.
R is the radius of the cycle wheel.
We know that the angular frequency is related to time period by the formula,
$\omega = \dfrac{{2\pi }}{T}$
So on putting the value of angular frequency in the equation of centripetal acceleration we have,
$\Rightarrow {a_c} = {(\dfrac{{2\pi }}{T})^2} \times 4 $
$\Rightarrow {a_c} = 4{\pi ^2} \times \dfrac{4}{4} $
$\Rightarrow {a_c} = 4{\pi ^2}{\text{m/}}{{\text{s}}^{\text{2}}}$
Hence the net acceleration is equal to centripetal acceleration in this case and hence the correct answer is option (A).

Note
There might be cases in which the tangential acceleration is also present. In such cases, we have to take the net resultant of both the accelerations ( i.e. both radial and tangential ) to find out the net acceleration. So in such cases the formula becomes ${a_{net}} = \sqrt {a_c^2 + a_r^2} $ will be applicable.