Question #36567

a disc and a ring of same mass are rolling down fritionless inclined plane with same kinetic energy .the ratio of velocity of disc to velocity of ring is ??????????????????????

Expert's answer

a disc and a ring of same mass are rolling down frittionless inclined plane with same kinetic energy. the ratio of velocity of disc to velocity of ring is ?

Solution:

Kinetic energies of the ring and disc are equal:


Edisc=EringE_{\text{disc}} = E_{\text{ring}}


The kinetic energy of rotational motion:


E=Jω22+mϑ22E = \frac{J \cdot \omega^2}{2} + \frac{m \cdot \vartheta^2}{2}


, where JJ - moment of inertia, ω=ϑr\omega = \frac{\vartheta}{r} - angular velocity, rr - radius, mm - mass.

Moment of inertia of the disc and the ring:


Jdisk=mrdisc22Edisc=mrdisc222ωdisc2+mϑdisc22=mrdisc24(ϑdiscrdisc)2+mϑdisc22=mϑdisc24+mϑdisc22=3mϑdisc22\begin{aligned} J_{\text{disk}} &= \frac{m r_{\text{disc}}^2}{2} \Rightarrow \\ E_{\text{disc}} &= \frac{m r_{\text{disc}}^2}{2 \cdot 2} \cdot \omega_{\text{disc}}^2 + \frac{m \cdot \vartheta_{\text{disc}}^2}{2} = \frac{m r_{\text{disc}}^2}{4} \left(\frac{\vartheta_{\text{disc}}}{r_{\text{disc}}}\right)^2 + \frac{m \cdot \vartheta_{\text{disc}}^2}{2} = \frac{m \vartheta_{\text{disc}}^2}{4} + \frac{m \cdot \vartheta_{\text{disc}}^2}{2} \\ &= \frac{3m \cdot \vartheta_{\text{disc}}^2}{2} \end{aligned}Jring=mr2Ering=mrring22ωring2+mϑring22=mrring24(ϑringrring)2+mϑring22==mϑring22+mϑring22=mϑring2\begin{aligned} J_{\text{ring}} &= m r^2 \Rightarrow \\ E_{\text{ring}} &= \frac{m r_{\text{ring}}^2}{2} \cdot \omega_{\text{ring}}^2 + \frac{m \cdot \vartheta_{\text{ring}}^2}{2} = \frac{m r_{\text{ring}}^2}{4} \left(\frac{\vartheta_{\text{ring}}}{r_{\text{ring}}}\right)^2 + \frac{m \cdot \vartheta_{\text{ring}}^2}{2} = \\ &= \frac{m \vartheta_{\text{ring}}^2}{2} + \frac{m \cdot \vartheta_{\text{ring}}^2}{2} = m \cdot \vartheta_{\text{ring}}^2 \end{aligned}


(3) and (2) in (1):


3mϑdisc22=mϑring2ϑdisc2ϑring2=23ϑdiscϑring=23=0.8\begin{aligned} \frac{3m \cdot \vartheta_{\text{disc}}^2}{2} &= m \cdot \vartheta_{\text{ring}}^2 \\ \frac{\vartheta_{\text{disc}}^2}{\vartheta_{\text{ring}}^2} &= \frac{2}{3} \\ \frac{\vartheta_{\text{disc}}}{\vartheta_{\text{ring}}} &= \sqrt{\frac{2}{3}} = 0.8 \end{aligned}


Answer: the ratio of velocity of disc to velocity of ring is 0.8.

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