Question

Ques no. 26 on
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Answer 1

Answer on Question #56549 - Physics - Mechanics - Relativity

(D) All the particles on the surface have the same linear speed.

5-26 A plank $P$ is placed on a solid cylinder $S$ , which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of the kinetic energy of $P$ to the kinetic energy of $S$ is:

(A) $1:1$

(B) $2:1$

(C) $8:3$

(D) $11:8$

Figure 5.92

Solution.

Let cylinder is moving with speed $V$ . The kinetic energy of $S(E_s)$ is consist of the energy of translatory motion $E_t$ and the energy of rotary motion $E_r$ .

$$E _ {s} = E _ {t} + E _ {r};$$ $$E _ {t} = \frac {m V ^ {2}}{2};$$ $$E _ {r} = \frac {I \omega^ {2}}{2},$$

Where $I$ is moment of inertia of cylinder (for solid cylinder $I = \frac{mR^2}{2}$ ), $\omega$ is angular velocity ( $\omega = \frac{V}{R}$ ). So

$$E _ {r} = \frac {1}{2} * \frac {m R ^ {2}}{2} * \left(\frac {V}{R}\right) ^ {2} = \frac {m V ^ {2}}{4};$$ $$E _ {s} = \frac {m V ^ {2}}{2} + \frac {m V ^ {2}}{4} = \frac {3 m V ^ {2}}{4}.$$

The speed the plank $\mathsf{P}(V_p)$ is equal to the speed of the top point cylinder which is twice more than speed of translatory motion. So $V_{p} = 2V$ .

$$E _ {p} = \frac {m V _ {p} ^ {2}}{2} = \frac {m (2 V) ^ {2}}{2} = 2 m V ^ {2};$$ $$\frac {E _ {p}}{E _ {s}} = \frac {2 m V ^ {2}}{\frac {3 m V ^ {2}}{4}} = \frac {8}{3}$$

Answer: C (8:3).

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