302 - Mechanics part 1 Questions Answers

Asked By: MITUL
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Use formula F(min)/mg = mg/F(max)

Now solve. Watch this video

Asked By: MITUL
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Asked By: SYED SAHIB AKTAR
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Asked By: MITUL
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Sir pls solve

Asked By: KANDUKURI ASHISH
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Impulse is Ft and Power = W/t F=mg and W = mgh Now calculate the time of motion  and get the answer

Sir please take this question as a Youtube video. Your explanation is great.
Asked By: AMAN KUMAR
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No horizontal force.

Draw a perpendicular from centre to horizontal plane. Then calculate horizontal distance. 

Now solve. 

If any problem, then inform

Sir please make a video explaining this problem solution as i love your explanation
Asked By: PRATEEK MOURYA
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According to given condition dθdt=vsin60a-(v+vcos60)t Now solve it for t then distance = vtsolve 

inform for any difficulty

A long smooth cylindrical pipe of radius r is tilted at an angle alpha to the horizontal. A small body at point A is pushed upwards along the inner surface of the pipe so that the direction of its initial velocity forms an angle phi with generatrix AB. Determine the minimum initial velocity V at which the body starts moving upwards without being separated from the surface of the pipe

Asked By: MAAZ
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At the front of the cylinder, see a tilted circle Now compare this problem with vertical circular motion. A is the top point of that path. We know that the  minimum velocity at top point tangential to path of  vertical circle should be gr. Here due to tilt of this circle effective gravity = gcosα so vsinφ =gcosαr now solve inform for any issue wait for video

Sir please give hint 
Asked By: KANDUKURI ASHISH
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According to first given condition velocity component of rain = u

let vertical component of rain = v

According to second condition 

velocity of rain relative to man 

=(nu-u)(-i^)+v(-j^)here direction of motion of man is i^ and  vertically upward direction is considered as j^ Now draw triangle for this vector and get cotθ solve it  inform for any issue 

Plz help sir. EDIT: Sir can you also tell how to solve part E?
Asked By: DARIUS
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Joshi sir comment

Updated:

Initially there is no external force. Centre of mass of the system accelerates when insect moves with acceleration.

Now suppose displacement of insect at t = t is x with respect to rod, and y is the acceleration of rod in downward direction with respect to ground. Thus acceleration of counterweight is y with respect to ground. NowxCM=[m(x-y)j^-(M-m)yj^+Myj^]/2M Now solve for x = l,

For dynamics, take friction upward in free body diagram of insect and downward in the free body diagram of rod.

Now solve

Inform for any issue.

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