# 223 - Mechanics part 1 Questions Answers

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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**Joshi sir comment**

$Accordingtogivencondition2r={{u}_{y}}^{2}/2g,\phantom{\rule{0ex}{0ex}}T=2{u}_{y}/g=2\sqrt{4gr}/g=4\sqrt{r/g}\phantom{\rule{0ex}{0ex}}Attoppointduetotouch,vecocityofthefrogbe{u}_{x}-2v.\phantom{\rule{0ex}{0ex}}R=[{u}_{x}{u}_{y}/g]+[{u}_{x}-2v){u}_{y}/g]=[{u}_{y}/g\left]\right(2{u}_{x}-2v)\phantom{\rule{0ex}{0ex}}=[2{u}_{y}/g\left]\right({u}_{x}-v)=T({u}_{x}-v)\phantom{\rule{0ex}{0ex}}Nowminimumpossiblevalueofvelocityoffrogattopto\phantom{\rule{0ex}{0ex}}crossthecylinderis\sqrt{gr}(Bytheconceptofverticalcircle)\phantom{\rule{0ex}{0ex}}nowsolvecarefullywithouterror$

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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**Joshi sir comment**

$Lettheaccelerationofhemisphereis{a}_{h}\phantom{\rule{0ex}{0ex}}N\mathrm{sin}\theta =M{a}_{h}\phantom{\rule{0ex}{0ex}}\Rightarrow N/m{a}_{h}=M/m\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}Herepseudoforceisconsideredduetoaccelerationofhemisphere$

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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**Asked By: AVRANIL SAHA**

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**Joshi sir comment**

$InfirstcaseF=2T=150\phantom{\rule{0ex}{0ex}}soT=75newton,\phantom{\rule{0ex}{0ex}}Nowinsecondcasesystemisfreetomove\phantom{\rule{0ex}{0ex}}accelerationofthesystema=F/20\phantom{\rule{0ex}{0ex}}for10kgrodF-2T=10a=10(F/20)=F/2\phantom{\rule{0ex}{0ex}}\Rightarrow F-(F/2)=2T\Rightarrow F=4T=300newton$

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

**Asked By: SHIV GOVIND**

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**Joshi sir comment**

$Momentofinertiaofacubeaboutalineperpendicularto\phantom{\rule{0ex}{0ex}}itsonefaceandpas\mathrm{sin}gthroughitscentreism{a}^{2}/6\phantom{\rule{0ex}{0ex}}soaboutdiagonalinertia={l}^{2}{I}_{x}+{m}^{2}{I}_{y}+{n}^{2}{I}_{z}\phantom{\rule{0ex}{0ex}}=({l}^{2}+{m}^{2}+{n}^{2}){I}_{x}(here{I}_{x}={I}_{y}={I}_{z})\phantom{\rule{0ex}{0ex}}={I}_{x}=m{a}^{2}/6\phantom{\rule{0ex}{0ex}}l,m,naredirection\mathrm{cos}inesofthreelinespas\mathrm{sin}gthrouge\phantom{\rule{0ex}{0ex}}centreofcubeandperpendiculartoeachother$

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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**Joshi sir comment**

Problem is interesting yet not complete. Complete it to get answer

thanks

**Asked By: DARIUS**

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**Joshi sir comment**

$Atageneralpoint\phantom{\rule{0ex}{0ex}}-2T\mathrm{sin}\theta =ma(\theta ismeasuredfromhorizontal)\phantom{\rule{0ex}{0ex}}\Rightarrow a=-2Mg\theta /m=-2Mgy/ml\phantom{\rule{0ex}{0ex}}comparewitha=-{\omega}^{2}y\phantom{\rule{0ex}{0ex}}\omega =\sqrt{\frac{2Mg}{ml}}\left(1\right)\phantom{\rule{0ex}{0ex}}Nowformaxdisplacementofsmallm\phantom{\rule{0ex}{0ex}}mg(h+y)=2Mgxandx={y}^{2}/2l\phantom{\rule{0ex}{0ex}}onsolvingthisquadraticeq.fory/lweget\phantom{\rule{0ex}{0ex}}\frac{y}{l}-\frac{m}{2M}=\sqrt{\frac{m}{M}\frac{h}{l}}\Rightarrow \frac{y}{l}=\sqrt{\frac{m}{M}\frac{h}{l}}(giventhat\frac{m}{M}\frac{h}{l})\phantom{\rule{0ex}{0ex}}sox=\frac{m}{2M}h\left(2\right)\phantom{\rule{0ex}{0ex}}so{v}_{max}=x\omega \phantom{\rule{0ex}{0ex}}nowsolve$

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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**Joshi sir comment**

$letuandvarethespeedsofboyandescalator\phantom{\rule{0ex}{0ex}}accordingtogivencondition\phantom{\rule{0ex}{0ex}}\left[1\right(u+v)T-2(u-v\left)T\right]{n}_{1}=\left[2\right(u+v)T-1(u-v\left)T\right]{n}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]T{n}_{1}=\left[2\right(u+v)-1(u-v\left)\right]T{n}_{2}\phantom{\rule{0ex}{0ex}}GiventhatT{n}_{1}=250andT{n}_{2}=50\phantom{\rule{0ex}{0ex}}\Rightarrow \left[1\right(u+v)-2(u-v\left)\right]250=\left[2\right(u+v)-1(u-v\left)\right]50\phantom{\rule{0ex}{0ex}}nowsolveitforv,uisgiven$

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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