223 - Mechanics part 1 Questions Answers

Plz help sir
Asked By: DARIUS
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A cylindrical pipe of radius r is rolling towards a frog sitting on the horizontal ground. Center of the pipe is moving with velocity v. To save itself ,the frog jumps off and passes over the pipe touching it only at top. Denoting air time of frog by T, horizontal range of the jump by R and acceleration due to gravity by g. Find T and R.

Asked By: MUDIT
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According to given condition 2r = uy2/2g, T= 2uy/g =24gr/g=4r/g At top point due to touch, vecocity of the frog be ux2v. R=[uxuy/g]+[ux2v)uy/g]=[uy/g](2ux2v) =[2uy/g](uxv) =T(uxv) Now minimum possible value of velocity of frog at top to  cross the cylinder is gr (By the concept of vertical circle) now solve carefully without error

A particle of mass m = 2kg is placed on the top of a hemisphere of M 4 kg. The hemisphere is placed on smooth ground. The particle is displaced gently. Then ratio of normal and pseudo force(seen from hemisphere frame) acting on particle when theta = 30 degree. Assume particle remain in contact with hemisphere.

Asked By: GUDDU KUMAR
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Let the acceleration of hemisphere is ah N sinθ = Mah N/mah = M/msinθ  Here pseudo force is considered due to acceleration of hemisphere 

A small ball is thrown from foot of a wall with minimum possible velocity to hit a bulb B on the ground at a distance L away from the wall.Find the expression for height h of shadow of the ball on the wall as a function of time t.Acceleration due to gravity is g.

Asked By: MUDIT
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How to solve this?
Asked By: AVRANIL SAHA
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In first case F = 2T = 150 so T = 75 newton,  Now in second case system is free to move acceleration of the system a = F/20 for 10 kg rod F-2T = 10a = 10(F/20) = F/2 F-(F/2)=2T F=4T = 300 newton 

what is the moment of inertia of a cube about its diagonal?

Asked By: SHIV GOVIND
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Moment of inertia of a cube about a line perpendicular to its one face and passing through its centre is ma2/6 so about diagonal inertia = l2Ix+m2Iy+n2Iz  = (l2+m2+n2)Ix    (here Ix=Iy=Iz) = Ix  = ma2/6 l,m,n are direction cosines of three lines passing througe centre of cube and perpendicular to each other

While making this paper, I stayed up till late at night working with candle light. I had two

candles of equal length L lit and placed at the same time, as shown in the figure. Immediately

thereafter I noticed that the shadow of the first candle on the left wall is not changing in

length, and the shadow of the second candle on the right wall is being shortened at a speed v.

(Take d1

= 10 cm, d2

= 20 cm, d3

= 30 cm, v = 1 mm/s, L = 20 cm.)

Asked By: AAYUSH MALAVIYA
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Problem is interesting yet not complete. Complete it to get answer 

thanks

Asked By: DARIUS
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At a general point -2Tsinθ = ma    (θ is measured from horizontal) a = -2Mgθ/m = -2Mgy/ml compare with a = -ω2y ω=2Mgml            (1) Now for max displacement of small m mg(h+y) = 2Mgx and x = y2/2l on solving this quadratic eq. for y/l we get  yl-m2M=mMhlyl=mMhl  (given that mM <<hl) so x = m2Mh       (2) so vmax = xω now solve

Two boys enter a running escalator at the ground floor of a shopping mall . The first boy repeatedly follows a cycle of p1=1 step up and then q1=2 steps down whereas the second boy repeatedly follows a cycle of p2=2 steps up and then q2=1 step down. Both of them move relative to escalator with a speed v=50cm/s . If the boys take t1=250s and t2=50s respectively to reach the first floor in complete number of cycles, how fast is the escalator running ? 

 

Asked By: JATIN KUMAR
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let u and v are the speeds of boy and escalator according to given condition  [1(u+v)T-2(u-v)T]n1=[2(u+v)T-1(u-v)T]n2  [1(u+v)-2(u-v)]Tn1=[2(u+v)-1(u-v)]Tn2 Given that Tn1=250 and Tn2=50 [1(u+v)-2(u-v)]250=[2(u+v)-1(u-v)]50 now solve it for v, u is given 

A particle projected from the ground passes two points , which are at height 12m and 18m above the ground and a distance d=10m apart . What could be the minimum speed of projection. Acceleration due to gravity is g=10m/s^2 . 

Asked By: JATIN KUMAR
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