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Question Number 214398 by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$$findaccelerationofwedgeA=?$$$$cylinderwithmassmandradiusr$$$$rollsalongthewedgewithmassM.$$

Answered by mr W last updated on 07/Dec/24

Commented by mr W last updated on 08/Dec/24

$$a=relativeaccelerationofm$$$$MA=N\sin \theta +f\cos \theta ...\left(i\right)$$$$-mA\sin \theta =N-mg\cos \theta ...\left(ii\right)$$$$m(a-A\cos \theta )=mg\sin \theta +f...\left(iii\right)$$$$\frac{{mr}^2}{2}\times \frac{a}{r}=-fr...\left(iv\right)$$$$\Rightarrow f=-\frac{ma}{2}$$$$ma-mA\cos \theta =mg\sin \theta -\frac{ma}{2}$$$$\Rightarrow a=\frac{2(g\sin \theta +A\cos \theta )}{3}$$$$\Rightarrow N=m(g\cos \theta -A\sin \theta )$$$$MA=m(g\cos \theta -A\sin \theta )\sin \theta -\frac{m}{2}\times \frac{2(g\sin \theta +A\cos \theta )}{3}\cos \theta$$$$3\frac{M}{m}A=2g\sin \theta \cos \theta -3A+2A{\cos }^2\theta$$$$\Rightarrow A=\frac{g\sin \theta \cos \theta }{\frac{3}{2}(\frac{M}{m}+1)-{\cos }^2\theta }$$

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