MCQ
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
- ✓$3 J$
- B$4 J$
- C$5 J$
- D$6 J$
✓
Answer
Correct option: A.
$3 J$
(a) : Total kinetic energy of the ring when it tolls without slipping,
$
K_{\text {ring }}=K_T+K_R=\frac{1}{2} m v^2+\frac{1}{2} I_r \omega^2
$
$\begin{aligned} & \quad=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \times \frac{v^2}{r^2} \quad\left(\because I_r=m r^2 \text { and } \omega=\frac{v}{r}\right) \\ & \quad=m v^2 \\ & \text { But } K_{\text {ring }}=4 J \text { (given) } \\ & \therefore \quad m v^2=4 J .....(i)\\ & \text { Similarly, } K_{\text {dise }}=\frac{1}{2} m v^2+\frac{1}{2} I_d \omega^2 \quad \text { } \\ & \quad=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{m r^2}{2} \times \frac{v^2}{r^2} \quad\left(\because I_d=\frac{m r^2}{2}\right) \\ & \quad=\frac{3}{4} m v^2=\frac{3}{4} \times 4 J =3 J \quad \text { (Using (i)) }\end{aligned}$
$
K_{\text {ring }}=K_T+K_R=\frac{1}{2} m v^2+\frac{1}{2} I_r \omega^2
$
$\begin{aligned} & \quad=\frac{1}{2} m v^2+\frac{1}{2} m r^2 \times \frac{v^2}{r^2} \quad\left(\because I_r=m r^2 \text { and } \omega=\frac{v}{r}\right) \\ & \quad=m v^2 \\ & \text { But } K_{\text {ring }}=4 J \text { (given) } \\ & \therefore \quad m v^2=4 J .....(i)\\ & \text { Similarly, } K_{\text {dise }}=\frac{1}{2} m v^2+\frac{1}{2} I_d \omega^2 \quad \text { } \\ & \quad=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{m r^2}{2} \times \frac{v^2}{r^2} \quad\left(\because I_d=\frac{m r^2}{2}\right) \\ & \quad=\frac{3}{4} m v^2=\frac{3}{4} \times 4 J =3 J \quad \text { (Using (i)) }\end{aligned}$
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