The spin-drier of a washing machine slows down uniformly from 880 rpm (revolutions per minute) to 260 rpm while making  50 revolutions.  Find the angular acceleration through these 50 revolutions.  Express your answer in rad/s2?

The answer is apparently -12.3359871531 revolutions. I got -13.418.

I did this by taking the difference of the rpms divided by 60min times 2pi to get -64.926. Then I took 60s/620rpm times 50 revolutions to get 4.849s. I divided the -62.926 by the 2.849s. Where am I going wrong?

2 Answers
A08
Mar 16, 2018

There is one to one correspondence between kinematic equations for linear motion and circular motion.

Writing the applicable kinematic expression in the instance case for a uniform angular acceleration #alpha# and inserting given values in SI units we get

#omega_f^2-omega_i^2=2alphatheta#
#=>(2pin_f)^2-(2pin_i)^2=2alpha(2pixx50)#
#=>(2pixx260/60)^2-(2pixx880/60)^2=2alpha(2pixx50)#
#=>alpha=((2pixx260/60)^2-(2pixx880/60)^2)/((2^2pixx50))#
#=>alpha=(2pi[(13/3)^2-(44/3)^2])/100#

Using the identity #[a^2-b^2=(a+b)(a-b)]#, we get

#alpha=(2pi(13/3+44/3)(13/3-44/3))/100#
#=>alpha=(2pi(57/3)(-31/3))/100#
#=>alpha=-(2pixx19xx31)/300#
#=>alpha=-12.3\ radcdot s^-2#, rounded to one decimal place.

Narad T.
Mar 16, 2018

The angular acceleration is #=-12.33rads^-1#

Explanation:

The initial angular velocity is #omega_0=880 xx2pi/60=92.15rads^-1#

The final angular velocity is #omega_1=260xx2pi/60=27.23rads^-1#

The angle is #theta=50xx2pi=100pirad#

Apply the equation of motion (rotation)

#omega_1^2=omega_0^2+2theta alpha#

The angular acceleration is

#alpha=(omega_1^2-omega_0^2)/(2theta)=(27.23^2-92.15^2)/(2*100pi)#

#=-12.33rads^-2#

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