Question

A wheel rolls purely on ground. If the velocity of a point 'P as shown on periphery of a body has a velocity equalto the velocity of the centre of mass of the body. Then theta = ?sm(1) 120(3) 60(2) 30°(4) 90°​

Answer 1

Value of $\bf{\theta}$ is $\bf{120^o}$

Explanation:

The wheel is purely rolling, meaning it rotates through the point of touching the ground around an axis.

The net velocity on the periphery at point P, relative to the wheel bottom, is:

$v = \omega r$

Where r is the distance from the end of the wheel to P

We have to use some geometry to find r.   We know from the Inscribed Angle Theorem that the angle inscribed is half the angle of the arc.   And from the theorem of Thales, we know that a right angle is an angle inscribed across a diameter.  

Thus,

$cos (\theta/2) = r / (2R)$

$r = 2R cos (\theta/2)$

then net velocity at point P,

$v = 2R\omega\;cos (\theta/2)$

Above velocity is equivalent to the velocity at the center of the circle, given by $R\omega$

$R\omega = 2R\omega\; cos (\theta/2)$

$1 = 2 cos (\theta/2)$

$cos (\theta/2) = 1/2$

$\theta/2 = 60^o$

$\theta= 120^o$

Thus, value of $\bf{\theta}$ is $\bf{120^o}$