Question

A uniform cylinder rolls down
from rest, on a track whose
vertical cross-section is a
parabola given by the equation
y = kx². If the surface is
rough from A to B due to
which the cylinder doesn't slip
but it is frictionless from B to
C, then the height of ascent of
cylinder towards C is
$a) \frac{y1}{3}$
$b) \frac{2y1}{3}$
$c) \frac{3y1}{2}$
$d) y1$
answer only if u know

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Answer 1

Answer:

this is the answer...

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Answer 2

Height of ascent of cylinder is $\frac{2}{3}y_1$

Therefore, option (b) is correct.

Explanation:

The body starts from rests and rolls without slipping

If after rolling down height $y_1$, its velocity is $v$ and angular velocity $\omega$

If radius of the cross section of cylinder is $R$ and its mass $M$

Then

Moment of inertia of cylinder

$I=\frac{1}{2}MR^2$

And

$Mgy_1=\frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$

$\implies 2Mgy_1=Mv^2+\frac{1}{2}MR^2\times(\frac{v}{R})^2$

$\implies 4gy_1=2v^2+v^2$

$\implies v^2=\frac{4}{3}gy_1$

After reaching the point B since the surface is smooth

Therefore, the body will not roll instead it will have only velocity $v$

If the height reached by the body is $h$

Then

$\frac{1}{2}Mv^2=Mgh$

$\implies \frac{1}{2}M\times\frac{4}{3}gy_1=Mgh$

$\implies \frac{2}{3}y_1=h$

or $h=\frac{2}{3}y_1$

Hope this answer is helpful.

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