Question

A wheel of radius r is rolling on a horizontal Surface At t=0 T is top most point on the wheel then magnitude of displacement of point T when wheel has completed one fourth rolling is

Answer 1

Answer:

$r\sqrt{(1+(\frac{\pi}{2}+1)^2}$

Explanation:

Given radius of the wheel = r

circumference of the ring = $2\pi r$

∵ In one complete revolution, the ring will cover a horizontal distance = $2\pi r$ m

∴ In 1/4 revolution, the horizontal distance covered = $\frac{1}{4}\times 2\pi r$ = $\frac{\pi r}{2}$

When the wheel starts rolling and completes 1/4 revolution the position of point T will be at T' as shown in figure

Here, $OT'=\frac{\pi r}{2}+r$

The displacement of point T will be TT'

In ΔTT'O

$(T'T)^2=(OT)^2+(OT')^2$

$\implies (T'T)^2 = r^2+(\frac{\pi r}{2}+r)^2$

$\implies (T'T)^2 = r^2(1+(\frac{\pi}{2}+1)^2$

$\implies T'T = r\sqrt{(1+(\frac{\pi}{2}+1)^2}$

Therefore, the displacement of point T is $r\sqrt{(1+(\frac{\pi}{2}+1)^2}$

Hope this helps.