Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is μ. Now the disc is pulled with a force F as shown in the figure.

A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between disc and table is \mu. Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Answers (1)

Let us assume a to be the linear acceleration and \alpha as angular acceleration.

F-f=Ma (In case of linear motion )

Torque\; to\; disc\; (\tau )=I_{D}\alpha

MI\; of\; disc\; is\; =I_{D}=\frac{1}{2}MR^{2}

f.R=\frac{1}{2}MR^{2}\left ( \frac{a}{R} \right )\; \; \; \; \; \; \; (a=R\alpha )

fR=\frac{1}{2}MRa

Ma=2f

F-f=2f

3f=F

f=\frac{F}{3}

\mu Mg=\frac{F}{3}

F=3\mu Mg;

This is the maximum force applied on the disc to roll on surface without slipping.

 

Posted by

infoexpert23

View full answer

Similar Questions

Latest Question