Q. A particle is released from the top of a fixed rough sphere. The particle leaves the surface of the sphere at `theta =` 60°. Find the work done by friction on the particle.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The figure is below. When the angle is `alpha < theta` the normal component of the weight G is

`Gn = G*cos(alpha) = m*g*cos(alpha)`

The friction force is

`Ff = mu*Gn =mu*m*g*cos(alpha)`

The length of arc circle between 0 and 60 degree (`pi/3` rad) is

`L =r*theta`

therefore the infinitesimal displacement along the circle is

`dL =r*d(alpha)` where angle `alpha` is measured in radians, and r is the radius of the sphere.

The total work done by the friction force is

`W = int_0^(pi/3)Ff*r*d(alpha) =int_0^(pi/3)mu*m*g*r*cos(alpha)d(alpha)=`

`=mu*m*g*r*sin(pi/3) =0.866*mu*m*g*r`

` `

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team

We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

  • 30,000+ book summaries
  • 20% study tools discount
  • Ad-free content
  • PDF downloads
  • 300,000+ answers
  • 5-star customer support
Start your 48-Hour Free Trial