# 42 - Rotational mechanics Questions Answers

**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**

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

A rotating ball hits a rough horizontal plane with a vertical velocity and angular velocity . Given that the coefficient of friction is and the vertical component of the velocity after the collision is , find:

a) the angular velocity after collision;

b) the impulsive ground reaction during the collision

**Asked By: UDDESH KUMAR SABAT**

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**Joshi sir comment**

on applying momentum impulse theorem along vertical

-mv+Ndt = mv/2 (1)

by angular momentum angular impulse theorem

$2m\omega {r}^{2}/5-\mu Ndtr=2m{\omega}^{,}{r}^{2}/5\left(2\right)$By eq. (2) get angular velocity.

impulsive ground reaction = $\sqrt{{\left(Ndt\right)}^{2}+{\left(\mu Ndt\right)}^{2}}$by using eq. (1), solve it

**What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination θ?**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

friction f = mgsinθ/[1+(mr^{2}/I)]

solve for sphere and compare to μmgcosθ

Find the moment of inertia of a uniform square plate of mass M and edge a about one of its diagonals.

(1) Ma^{2 }/ 12

(2) 2/3 x Ma^{2}

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

let the moment of inertia about a diagonal = I

then by perpendicular axis theorem 2I = 1/6 Ma^{2}

now calculate I

**Asked By: MANISH KUMAR SAHU**

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**Joshi sir comment**

let the angular velocities are ω_{1}, ω_{2}

then angular acceleration of one w. r. t. other = ω_{1} ω_{2}

**A ball of mass 1 kg is projected with a velocity of 20√2 m/s from the origin of an xy coordinates axis system at an angle 45° with x-axis (horizontal). The angular momentum of the ball about the point of projection after 2s of projection is [take g=10 m/s ^{2}] (y-axis is taken as vertical)**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

initial vertical and horizontal velocities are 20 m/s and 20 m/s

after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s

and y and x displacements are 20 m and 40 m

so about point of projection angular momentum after 2 sec.

= mv_{x}y - mv_{y}x = 1*20*20 - 1*0*40 = 400

**A disc of mass 3kg rolls down an inclined plane of height 5 m. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

mgh = 1/2 mv^{2} + 1/2 Iω^{2}

I for the disc about centre = 1/2 MR^{2 }

and for pure rolling v = Rω

calculate v then 1/2 mv^{2}

**A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

for translational motion

0-μMg = Ma so a = -μg

so after time t velocity v = u-μgt (1)

for rotational motion

about centre, μMgr = Iα = 2Mr^{2}α/5

so α = 5μg/2r

so ω = 0 + αt

so ω = 5μgt/2r (2)

for pure rolling

v = rω implies that u-μgt = r5μgt/2r

implies that u = 7μgt/2

so t = 2u/7μg