Question

An elevator is going up with an acceleration $2\,m/{s^2}$ . If the radius of the wheel attached to the elevator is $0.1$ m, then find out the number of revolutions in $t = 10s$.

Answer 1

You can easily solve this question if you understand that the rope will circumference the wheel attached to the elevator. This problem can be solved by using equations of motion for constant acceleration. The relationship between the number of revolutions and the angular acceleration will be used in solving the problem.

Formula used:

Angular Displacement = $\theta$= ${\omega_0}{t}+\dfrac {1}{2} {\alpha}{t^2}$

Where,

${\omega_0}$- Initial velocity

${\alpha}$- The angular acceleration

$t$- The total time taken

Complete solution:

We will be trying to solve the question exactly as told in the hint section of the solution to this question. First, we will process the information about rope getting wrapped up along the circumference of the wheel, then we will use this information to find a relation between the angular displacement and the number of revolutions made by the wheel in that time interval.

Given that,

The acceleration of the elevator=$a$= $2 m/{s^2}$

The radius of the wheel=$r$=$0.1\;m$

Now, we are asked to calculate the number of revolutions in $t=10s$. Let the angular displacement of the wheel be $\theta$. The angular displacement of the wheel will be $2\pi$ times the number of revolutions $n$. The angular displacement of the wheel can be calculated using the following mathematical equation:

${\theta}= 2{\pi}\times n= {\omega_0}{t}+\dfrac {1}{2} {\alpha}{t^2}$ ….. (1)

Since initial velocity is zero, we get:

$2{\pi}\times n= \dfrac {1}{2} {\alpha}{t^2}$ ….(2)

Using the relation between the angular acceleration and the linear acceleration, we get:

$a= \alpha r$

$\alpha = \dfrac {a}{r}$ …..(3)

Substituting equation (3) in (2) we get:

$\Rightarrow 2{\pi}\times n= \dfrac {1}{2} {\dfrac {a}{r}}{t^2}$

$\Rightarrow n= {\dfrac {1}{2}} \left ({\dfrac {a}{r}}\right){\dfrac {t^2}{2\pi}}$

$\Rightarrow n= {\dfrac {1}{2}} \left ({\dfrac {2} {0.1}}\right) {\dfrac {100}{2\pi}}$

$\Rightarrow n=159$

Therefore, the wheel makes 159 revolutions in $t=10\;s$.

Note:**** Many students forget to consider the fact that the rope is getting wrapped up along the circumference of the wheel and get stuck in the question. Another important point in the solution of the question is the fact that initial velocity of the elevator is to be taken as zero since nothing about it has been given to us by the question.