Solveeit Logo

Question

Question: The spin dryer of a washing machine rotating at 15 r.p.s. slows down to 5 r.p.s. after making 50 rev...

The spin dryer of a washing machine rotating at 15 r.p.s. slows down to 5 r.p.s. after making 50 revolutions. Find its angular acceleration.

Explanation

Solution

Hint: One complete revolution is equal to an angle of 2π\pi . Convert the angular velocities in the units of per second. Then use the equation 2αθ=ω22ω122\alpha \theta =\omega _{2}^{2}-\omega _{1}^{2} to find the angular acceleration of the spin dryer.

Formula used:
1 r.p.s = 2πs1\pi {{s}^{-1}}
2αθ=ω22ω122\alpha \theta =\omega _{2}^{2}-\omega _{1}^{2}

Complete step by step answer:
The angular velocity of a rotating or a revolving body is the change in angle of rotation in one unit of time. One of the units of angular velocity is r.p.s.
Let us understand the meaning of r.p.s. The full form of r.p.s is revolution per second. It is a unit of revolution. It tells us how many rotations or revolutions the body completes in one second of time. 1 complete revolution is when the revolving body rotates by an angle of 2π\pi . Therefore, 1 r.p.s = 2πs1\pi {{s}^{-1}}.
Therefore,
15 r.p.s = 15×2πs1=30πs115\times 2\pi {{s}^{-1}}=30\pi {{s}^{-1}}.
And
5 r.p.s = 5×2πs1=10πs15\times 2\pi {{s}^{-1}}=10\pi {{s}^{-1}}.
The rate of change of angular velocity with time is called angular acceleration. The SI unit of angular acceleration is per second square (s2{{s}^{-2}}).
Suppose a rotating body, changes its angular velocity from ω1{{\omega }_{1}} to ω2{{\omega }_{2}} and during this it rotates for an angle θ\theta . Then the relation between the angular velocities (ω1{{\omega }_{1}} and ω2{{\omega }_{2}}), angle of rotation (θ\theta ) and its angular acceleration (α\alpha ) is given by,
2αθ=ω22ω122\alpha \theta =\omega _{2}^{2}-\omega _{1}^{2} …… (i).
Let us consider the clockwise direction as positive direction and the anticlockwise as negative direction.
It is given that the angular velocity of the spin dyer changes from 15 r.p.s to 5 r.p.s.
This means ω1=15rps=30πs1{{\omega }_{1}}=15rps=30\pi {{s}^{-1}} and ω2=5rps=10πs1{{\omega }_{2}}=5rps=10\pi {{s}^{-1}}.
It is said that the dyer completes 50 revolutions.
Hence, θ=50revolutions=50×2π=100π\theta =50revolutions=50\times 2\pi =100\pi .
Substitute all the known values in equation (i).
2α(100π)=(10π)2(30π)2\Rightarrow 2\alpha (100\pi )={{\left( 10\pi \right)}^{2}}-{{\left( 30\pi \right)}^{2}}
200πα=100π2900π2\Rightarrow 200\pi \alpha =100{{\pi }^{2}}-900{{\pi }^{2}}
200πα=800π2\Rightarrow 200\pi \alpha =-800{{\pi }^{2}}
α=800π2200π=4πs2\Rightarrow \alpha =\dfrac{-800{{\pi }^{2}}}{200\pi }=-4\pi {{s}^{-2}}.
The angular acceleration is negative because the dyer is slowing down.

Note: Rotation mechanics can be correlated with translation mechanics (motion in straight lines). We say that rotation motion is analogous to translation motion.
The quantity displacement is analogous to the angle of rotation.
Velocity is analogous to angular velocity.
Acceleration is analogous to angular acceleration.
Therefore, the kinematic equations for translation motion are also analogous to rotation motion.
v=u+atv=u+at is analogous to ω2=ω1+αt{{\omega }_{2}}={{\omega }_{1}}+\alpha t
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}} is analogous to θ=ω1t+12αt2\theta ={{\omega }_{1}}t+\dfrac{1}{2}\alpha {{t}^{2}}
2as=v2u22as={{v}^{2}}-{{u}^{2}} is analogous to 2αθ=ω22ω122\alpha \theta =\omega _{2}^{2}-\omega _{1}^{2}.
Remember that θ\theta , ω\omega , α\alpha are vectors. Hence, you must take care of the direction of the quantities.