Question

A mass is attached to one end of a mass less string (length $l$), the other end of which is attached to a fixed support O. The mass swings around in a vertical circle as shown in fig. The mass has the minimum speed (u = $\sqrt{5gl})$ necessary at the lowermost point of the circle to keep the string from going slack. If you cut the string, when it is making angle $\theta$ (as shown in figure) with vertical line through centre O of the circle, the resulting projectile motion of the mass has its maximum height located directly above the center of the circle.

Find the value of $10\cos \theta :$

• A. $5$
• B. $19$
• C. $4$
• D. $22$

Answer 1

Answer: (A)

$5$

$velocityatthepointwherethestringwascut\frac{1}{2}m\left(5gl\right)=\frac{1}{2}m{v}^2+mgl(1+cos\theta )v=\sqrt{2gl(\frac{3}{2}-cos\theta )}R=\frac{v^{2}sin2\theta }{g}=2lsin\theta \Rightarrow 2gl(\frac{3}{2}-cos\theta )sin2\theta =2glsin\theta \Rightarrow 2(\frac{3}{2}-cos\theta )cos\theta =1\Rightarrow 2{cos}^2\theta -3cos\theta +1=0\Rightarrow cos\theta =\frac{1}{2}or1\Rightarrow 10cos\theta =5$