Question

A ring and a disc roll on the horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is $4\, J$ then total kinetic energy of the disc is

• A. $3\, J$
• B. $4\, J$
• C. $5\, J$
• D. $6\, J$

Answer 1

The Correct Option is A

Answer (A) \(3\, J\)

Answer 2

Explanation:

\(\text{Kinetic Energy of the rolling body is given by:}\)

\(K.E.=\frac{1}{2} mv^{2}(1+\frac{k^{2}}{r^{2}})\)

For a ring \(\frac{k^{2}}{r^{2}}=1\)

\(\therefore KE_{R}= \frac{1}{2}mv^{2}(1+1)=mv^{2}\)

For a disc \(\frac{k^{2}}{r^{2}}=\frac{1}{2}\)

\(\therefore KE_{R}= \frac{1}{2}mv^{2}(1+1)=\frac{3}{4}mv^{2}\)

\(KE_{R}= mv^{2}=4J\)

\(KE_{D}= \frac{3}{4}mv^{2}=\frac{3}{4}\times 4 =3J\)

Answer 3

The correct option is (A)

The total kinetic energy of the body \(\text{K.E}_{total}​ = (\frac{1}{2}​mv^{2})_{translational​}+(\frac{1​}{2}Iw^{2})_{rotational​}\)

Total kinetic energy of the ring \(\text{K.E}_{ring}​=\frac{1}{2}​mv^{2}+\frac{1}{2}​(Mr^{2})w^{2}\)

\(\text{K.E}_{ring}​=\frac{1}{2}​mv^{2}+\frac{1}{2}​Mv^{2}=Mv^{2}\)       (for pure rolling \(v=rw\)\(\))

\(⟹ mv^{2}=4\)

Total kinetic energy of the disc \(\text{K.E}_{disc}​=\frac{1}{2}​mv^{2}+\frac{1}{2}​(\frac{1}{2}Mv^{2})w^{2}\)

\(\text{K.E}_{disc}​=\frac{1}{2}​mv^{2}+\frac{1}{4}​Mv^{2}=\frac{3}{4}Mv^{2}\)      (for pure rolling \(v=rw\))

\(⟹ \text{K.E}_{disc}​=43​\times4=3J\)