Question

A spool of mass $M=3kg$ and radius $R=20cm$ has an axle of radius $r=10cm$ around which a string is wrapped. The moment of inertia about an axis perpendicular to the plane of the spool and passing through the centre is $\frac{M{R}^2}{2}$ . If the coefficient of friction between the surface and the spool is $0.4$ then the maximum tension which can be applied to the string for which the spool doesn't slip, is $\left[g=10m{s}^{-2}\right]$

• A. $24N$
• B. $18N$
• C. $9N$
• D. $36N$

Answer 1

Answer: (B)

Writing the torque about the instantaneous axis of rotation we get,
$\tau =T\times \left(R-r\right)=\left(\frac{M{R}^2}{2}+M{R}^2\right)\alpha$
$\Rightarrow T=\left(\frac{3M{R}^2}{2\left(R-r\right)}\right)\frac{a}{R}$
Also,
$T-\mu Mg=Ma$
Substituting the values and solving for $T$ we get,
$T=18N$