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JEE Advanced 2015Paper 1
Exam
JEE Advanced
Year
2015
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2015 · Paper 1 · Physics · Rotational Motion

Two identical uniform discs roll without slipping on two different...

Two identical uniform discs roll without slipping on two different surfaces ABandCDAB and CD (see figure) starting at AandCA and C with linear speeds υ1andυ2{υ}_{1} and {υ}_{2} , respectively, and always remain in contact with the surfaces. If they reach BandDB and D with the same linear speed and υ1=3{υ}_{1} = 3 m/s, then υ2{υ}_{2} in m/s is \left(g = 10\dfrac{m}{{s}^{2}}\right)
Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C

Correct answer

7

Step-by-step solution

K.Ed=K.ET+K.ERotK . {E}_{d} = K . {E}_{T} + K . {E}_{Rot}
Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C
=12mv2+12Icmw2= \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}{I}_{cm}{w}^{2}
=12mv2+12(mr22)(vr)2= \dfrac{1}{2}m{v}^{2} + \dfrac{1}{2}\left(\dfrac{m{r}^{2}}{2}\right){\left(\dfrac{v}{r}\right)}^{2}
=34mv2= \dfrac{3}{4}m{v}^{2}
KA+Wg=KB=34mv12+mg×30=KB{K}_{A} + {W}_{g} = {K}_{B} = \dfrac{3}{4}{mv}_{1}^{2} + mg \times 30 = {K}_{B}
KC+Wg=KD=34mv22+mg×27=KD{K}_{C} + {W}_{g} = {K}_{D} = \dfrac{3}{4}m{v}_{2}^{2} + mg \times 27 = {K}_{D}
KB=KD{K}_{B} = {K}_{D} (Conversation of energy)
⇒34m(3)2−34mv22+3mg=0⇒ \dfrac{3}{4}m{\left(3\right)}^{2} - \dfrac{3}{4}m{v}_{2}^{2} + 3mg = 0
⇒v22=49⇒ {v}_{2}^{2} = 49
⇒v=7⇒ v = 7

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