A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2 - \sqrt{3}) / \sqrt{10} s$ , then the height of the top of the inclined plane, in meters, is ____________.
Take $g = 10 m / s^{2}$ .

Question

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2 - \sqrt{3}) / \sqrt{10} s$ , then the height of the top of the inclined plane, in meters, is ____________.

Take $g = 10 m / s^{2}$ .

Infinity Learn · Accepted Answer

$$ac=gsin$$⁡θ$$1+ICMR2$$ aring $$=gsin$$⁡θ$$2$$,adisc $$=2gsin$$⁡θ$$3$$ For ring hsin⁡θ$$=12gsin$$⁡θ$$2t12$$⇒$$t1=4hgsin2$$⁡θ$$=16h3g$$ For disc, hsin⁡θ$$=122gsin$$⁡θ$$3t22$$⇒$$t2=3hgsin2$$⁡θ$$=4hg$$ Given :$$t2$$−$$t1=16h3g$$−$$4hg=2$$−$$310$$⇒$$h43$$−$$2=2$$−$$3$$⇒$$h=(2$$−$$3)3(4$$−$$23)=32$$⇒$$h=0.75m$$

Rolling motion

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Rolling motion

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ac=gsin⁡θ1+ICMR2 aring =gsin⁡θ2,adisc =2gsin⁡θ3 For ring hsin⁡θ=12gsin⁡θ2t12⇒t1=4hgsin2⁡θ=16h3g For disc, hsin⁡θ=122gsin⁡θ3t22⇒t2=3hgsin2⁡θ=4hg Given :t2−t1=16h3g−4hg=2−310⇒h43−2=2−3⇒h=(2−3)3(4−23)=32⇒h=0.75m

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