Question

A circular disc of mass M and radius r is set rolling on a table . If it's angular velocity , show that its total kinetic energy is given by 3/4Mv^2 , where v is its linear velocity . M.I. of circular disc =1/2Mr^2.

Answer 1

Answer:

give any choice for this question

Answer 2

Answer:

$K.E.=\frac{3Mv^{2} }{4}$

Explanation:

Given A circular disc of mass M and radius r is set rolling on a table.

since during rolling it performs linear motion as well as rotational motion.

So,$Total K.E.=K.E_{linear\ motion} +K.E._{Rotation\ motion}$...........(1)

$K.E._{Linear motion} =\frac{mv^{2} }{2}$..............(2)

$K.E._{Rotation\ motion} =\frac{I\varpi^{2} }{2}$..............(3)

for a circular disc $I=\frac{Mr^{2} }{2}$.............(4)

we know that

$v=r\varpi$........................(5)

using (3),(4) and (5) we get

$K.E._{Rotation\ motion} =\frac{Mv^{2} }{4}$.......................(6)

using (1),(2) and (6) we get

$Total K.E.=\frac{Mv^{2} }{2} +\frac{Mv^{2} }{4}$

So, $K.E.=\frac{3Mv^{2} }{4}$ Hence proved.