Question

With the assumption of no slipping determine the mass $m$ of the block which must be placed on the top of a $5\text{kg}$ cart in order that the time period of oscillation is $2\text{sec}$. What is the minimum coefficient of static friction ${\mu }_s,$ for which the block will not slip relative to the cart, if the cart is displaced $300\text{mm},$ from the equilibrium position and released ?

• A. $m=2.08\text{kg},{\mu }_s\ge 0.402$
• B. $m=1.079\text{kg},{\mu }_s\le 0.302$
• C. $m=2.08\text{kg},{\mu }_s\le 0.402$
• D. $m=1.079\text{kg},{\mu }_s\ge 0.302$

Answer 1

The correct option is D $m=1.079\text{kg},{\mu }_s\ge 0.302$

As we know that
$T=2\pi \sqrt{\frac{5+m}{60}}$
$2=2\pi \sqrt{\frac{5+m}{60}}$
$\Rightarrow m=1.079\text{kg}$
For maximum acceleration of SHM.
$a_{max}={\omega }^{2}A$
So, maximum force on mass 'm' is $m{a}_{max}=m{\omega }^{2}A$. and this $m{\omega }^{2}A$ is provided by friction force only
So ${\mu }_{s}mg\ge m{\omega }^{2}A$
${\mu }_s\ge \frac{{\omega }^{2}A}{g}$
${\mu }_s\ge {\left(\frac{2\pi }{T}\right)}^2\frac{A}{g}$
$\Rightarrow {\mu }_s\ge {\left(\frac{2\pi }{2}\right)}^2\times \frac{0.3}{9.8}$
or
${\mu }_s\ge 0.302$