Solveeit Logo

Question

Question: If a cycle wheel of radius \(0.4{\text{m}}\) completes one revolution in \(2\) seconds then accelera...

If a cycle wheel of radius 0.4m0.4{\text{m}} completes one revolution in 22 seconds then acceleration of the cycle is
A. 0.4πm/s20.4\pi \,{\text{m/}}{{\text{s}}^{\text{2}}}
B.0.4π2m/s20.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}
C. π20.4m/s2\dfrac{{{\pi ^2}}}{{0.4}}\,{\text{m/}}{{\text{s}}^{\text{2}}}
D.0.4π2m/s2\dfrac{{0.4}}{{{\pi ^2}}}\,{\text{m/}}{{\text{s}}^{\text{2}}}

Explanation

Solution

In this question, first define angular velocity, acceleration, time period and then state the relationship among them. Then calculate the angular velocity and further use it to obtain the value of acceleration of the cycle wheel.

Complete step by step solution:
As we know that angular velocity is the rate of change of angular displacement. For an object rotating about an axis every point of the object will have the same angular velocity. It is denoted by ww.

In circular motion angular acceleration occurs when the rate of angular velocity changes. It is denoted as α\alpha .

We also know that when acceleration of a body is expressed as aa then it has a relationship with angular velocity. The relationship is expressed as

a=w2r \Rightarrow a = {w^2}r

Where ww is the angular velocity and rr is the radius of the circle.

Time period is known as the time taken for completing a complete revolution. It is denoted as TT.

We have given the following data
Radius of cycle wheel =r=0.4m = r = 0.4{\text{m}}

Time for one revolution is 22 seconds that is time period T=2T = 2 seconds.

We know that relation between angular velocity and time period is expressed as
w=2πT\Rightarrow w = \dfrac{{2\pi }}{T}
So here we can get the value of angular velocity as,
w=2π2=πrad/s\Rightarrow w = \dfrac{{2\pi }}{2} = \pi {\text{rad/s}}.

Now we will calculate the acceleration of the wheel as
a=w2r\Rightarrow a = {w^2}r

Now, we substitute the values in the above equation as,
a=π2×0.4m/s2\Rightarrow a = {\pi ^2} \times 0.4{\text{m/}}{{\text{s}}^{\text{2}}}

After simplification we get,
a=0.4π2m/s2\therefore a = 0.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}
Thus, we get the acceleration of the cycle wheel as 0.4π2m/s20.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}.

So, option(B) is the correct option.

Note: As we know that while solving the question put correct units in the equations. Also note that while revolutionizing the cycle wheel the center of the wheel follows translation and the rim of the wheel exhibits rotary motion.