Question

A small sphere of mass ‘m’ is tied to the top of the smooth inclined plane with the help of a string of length ‘L’. The string and the inclined plane makes an angle ‘$\theta$’ with the horizontal. The inclined plane is rotated with a constant angular velocity ‘$\omega$’ about the vertical axis passing through the end of the string fixed to the plane as shown in the figure.


If $\theta ={37}^{\circ },\omega =2rad/s,L=2.45m,$ then the ratio between the tension in the string and the normal reaction between the sphere and the plane is: [consider sin ${37}^{\circ }=\frac{3}{5}andg=9.8m/s^2]$

• A. $\frac{8}{31}$
• B. $\frac{16}{31}$
• C. $\frac{31}{8}$
• D. $\frac{31}{16}$

Answer 1

The correct option is C $\frac{31}{8}$


Along vertical axis,
$Tsin\theta +Ncos\theta =mg..........\left(i\right)$
Along radial axis,
$Tcos\theta -Nsin\theta =m{\omega }^2\left(Lcos\theta \right)$........(ii)
From (i) and (ii), we get
$N=mgcos\theta -m{\omega }^{2}Lcos\theta sin\theta$
$\omega =2rad/s,L=2.45m{\omega }^{2}L=4\times 2.45=9.8=g$
Hence, $\frac{T}{N}=\frac{mgsin\theta +m{\omega }^{2}Lco{s}^2\theta }{mgcos\theta -m{\omega }^{2}Lcos\theta sin\theta }$
$=\frac{31}{8}$