Question

Find the total acceleration of the sphere as a function of $\theta$, the angle of deflection of the thread from the vertical.

• A. $w=g\sqrt{1+2{\cos }^2\theta }$
• B. $w=g\sqrt{2+3{\cos }^2\theta }$
• C. $w=g\sqrt{1+3{\cos }^2\theta }$
• D. $w=g\sqrt{3+2{\cos }^2\theta }$

Answer 1

The correct option is C $w=g\sqrt{1+3{\cos }^2\theta }$

Let the thread make the angle $\theta$ with the vertical at time $t$ .

Using the conservation of mechanical energy, Gain in Kinetic energy $=$ Loss in potential energy.

thus, $\frac{1}{2}m{v}^2-0=mgL\cos \theta \Rightarrow v^2=2gL\cos \theta ....\left(1\right)$

Now the centripetal acceleration at this moment ,$a_n=\frac{v^2}{L}=2g\cos \theta$

The tangential acceleration at this moment, $a_t=g\sin \theta$

Thus the total acceleration ,$w=\sqrt{a_n^2+a_t^2}=\sqrt{4g^2{\cos }^2\theta +g^2{\sin }^2\theta }=g\sqrt{4{\cos }^2\theta +(1-{\cos }^2\theta )}=g\sqrt{3{\cos }^2\theta +1}$