Question

A cycle wheel of radius 0.4 m completes one revolution in one second then the acceleration of a point on the cycle wheel will be
A. \[0.8m/s\]
B. \[0.4m/s\]
C. \[1.6{\pi ^2}m/{s^2}\]
D. \[0.4{{\pi} ^2}m/{s^2}\]

Answer 1

Hint: To move the body in a circular path with uniform circular motion there needs a force which should balance the radially outward centrifugal force so that the body remains in the circular path. The radially inward force is called the centripetal force acting along the radius of the circular path.

Formula used:
\[{F_c} = m{\omega ^2}r\]
Here \[{F_c}\] is the centripetal force on a body of mass m moving in a circular path of radius r with angular velocity \[\omega \].
\[\omega = \dfrac{\theta }{t}\]
Here \[\omega \] is the angular velocity, \[\theta \] is the angular displacement in time t.
\[{F_{net}} = ma\]
Here \[{F_{net}}\] is the net force acting on the body of mass m and acceleration a.

Complete step by step solution:
The radius of the cycle wheel is given as 0.4 m
\[r = 0.4m\]
The cycle wheel completes the revolution in one second.
The angular displacement in one complete revolution is \[2\pi \] radian in 1 second.

Using the formula of angular velocity, the angular velocity of the cycle wheel is,
\[\omega = \dfrac{{2\pi }}{1}rad/s\]
\[\Rightarrow \omega = 2\pi rad/s\]
So, the centripetal acceleration will be,
\[{a_c} = \dfrac{{{F_c}}}{m}\]
\[\Rightarrow {a_c} = \dfrac{{m{{\left( {2\pi } \right)}^2} \times 0.4}}{m}m/{s^2}\]
\[\Rightarrow {a_c} = \left( {1.6{\pi ^2}} \right)m/{s^2}\]

As there is no tangential force acting, so the tangential acceleration is zero.
\[{a_t} = \dfrac{{{F_t}}}{m}\]
\[\Rightarrow {a_t} = 0\,m/{s^2}\]
The tangential acceleration is along the tangent to the circular path and the centripetal acceleration is along the line joining to the center of the circular path. Hence, the angle between the centripetal acceleration and the tangential acceleration is 90°.

The net acceleration will be the magnitude of the resultant of the tangential acceleration and the centripetal acceleration. So, the net acceleration at the point on the circular path is,
\[a = \sqrt {{{\left( {{a_c}} \right)}^2} + {{\left( {{a_t}} \right)}^2}} \]
\[\Rightarrow a = \sqrt {\left( {1.6{\pi ^2}} \right) + {{\left( 0 \right)}^2}} m/{s^2}\]
\[\therefore a = 1.6{\pi ^2}m/{s^2}\]
Hence, the acceleration at the point on the circular path is \[1.6{\pi ^2}m/{s^2}\].

Therefore, the correct option is C.

Note: The direction of acceleration for uniform circular motion is towards the center of the circular path and for accelerated circular motion, the direction of acceleration is along the resultant force of the tangential force and the centripetal force.