Question

A ring of radius 20 cm and mass 5 kg rolls on a smooth horizontal surface at the rate of 2 m/s . Rotational kinetic energy of ring is *​

Answer 1

Answer:

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Explanation:

Given

Radius of ring = 20cm = 0.2m

Mass of ring = 5kg

It rolls on a smooth surface at the rate = 2m/s

{\underline{\sf{To\:Find}}}

ToFind

Rotational kinetic energy of a ring

{\underline{\sf{Solution}}}

Solution

Let consider that ring is rotated about an Axis pperpendicular to plane passing through the centre of mass.

⇒ I = MR²

We know that Kinteic energy of a rolling body is :

\sf{K.E.=\dfrac{1}{2}mv{}^{2}+\dfrac{1}{2}I\omega}K.E.=

2

1

mv

2

+

2

1

Iω

\begin{lgathered}\sf{\implies \frac{1}{2} \times 5 \times {2}^{2} + \frac{1}{2} \times 5 \times {0.2}^{2} \times \frac{{2}^{2}}{{0.2}^{2}}}\\\end{lgathered}

⟹

2

1

×5×2

2

+

2

1

×5×0.2

2

×

0.2

2

2

2

\begin{lgathered}\sf{\implies 10 + \frac{1}{2} 5 \times 0.04 \times \frac{400}{4} }\\\end{lgathered}

⟹10+

2

1

5×0.04×

4

400

\begin{lgathered}\sf{\implies 10 + 5 \times 0.02 \times 100}\\\end{lgathered}

⟹10+5×0.02×100

\begin{lgathered}{\boxed{\sf{\implies 10 + 10 = 20 \; J}}}\\\end{lgathered}

⟹10+10=20J

More About rolling motion

When a body performs translatory motion as well as rotatory motion then it is known as rolling motion.