Question

​ A hoop rolls on a horizontal ground without slipping with linear speed v. Speed of particle P on the circumference of the hoop at angle θ is 1 2 vsinθ 3 4 vcosθ ​

Answer 1

the component along the motion is -v sin theta
The speed of point P wrt round is
v - v sin theta = v (1-sin theta)

Answer 2

The speed of particle on the circumference is $2 V \sin \left(\frac{\theta}{2}\right)$.

Given:

Hoop is rolling without slipping.

Solution:

The point P on the circumference at an angle theta will have velocity of vp which will have the distance from center to the point P multiplied with the angular velocity, due to the circular motion of hoop while rolling.  

$Vp = O P \times \omega$

Where,

$\omega = Angular\ Velocity$

$Vp = 2 R \sin \left(\frac{\theta}{2}\right) \times \omega$

We know that,

$\omega=\frac{V}{R} \Rightarrow V=\omega R$

$V p=2 V \sin \left(\frac{\theta}{2}\right)$.