42 - Rotational mechanics Questions Answers

How to solve this?
Asked By: AVRANIL SAHA
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Joshi sir comment

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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Joshi sir comment

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

Pls give Solution
Asked By: PRIYANSHU SINGH
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Joshi sir comment

A rotating ball hits a rough horizontal plane with a vertical velocity v and angular velocity w. Given that the coefficient of friction is u and the vertical component of the velocity after the collision is v/2, 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

2mωr2/5 - μNdtr = 2mω,r2/5       (2)By eq. (2) get angular velocity.

impulsive ground reaction = (Ndt)2+(μNdt)2by 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+(mr2/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/ 12

(2) 2/3 x Ma2

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 Ma2

now calculate I

Two sphere are rotating about their own axis if axis of rotation of these axis are perpendicular to each other than angular acceleration of one with respect to other
Asked By: MANISH KUMAR SAHU
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Joshi sir comment

let the angular velocities are ω1 ω2

then angular acceleration of one w. r. t. other =  ω1 ω2

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/s2] (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.

= mvxy - mvyx = 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 mv2 + 1/2 Iω2

I for the disc about centre = 1/2 MR

and for pure rolling v = Rω

calculate v then 1/2 mv2

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 = 2Mr2α/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 

now put t in (1) for getting v

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