Rotation - Quickies Homework Statement Solutions

[cupid.callin]

Homework Statement

The Attempt at a Solution

#1
I guess no. coz considering anybody from a point inside or not it ... its in pure rotation and for a point outside it ... it wil have both translational and rotational motion

#2
i don't know what they are asking .. if question is that can we write
A/a = v/w as (A+a) / (A-a) = (v+w) / (v-w) they yes. but won't that be a maths question

#3
Dont know ... some help please

#4
Dont know this either. yes i think that net torque will be zero but don't know how the it becomes horizontal

#5
My attempt:

I (moment of inertia) along BY is less .. so kinetic energy along BY is less and thus its easier to rotate along BY


#6
If ice melts ... radius increase ... as angular momentum is constant, I inc. and thus w dec ... so length of day-night inc
right?

#7
I guess solid sphere ... not just guess but worked out something ... someone just tell me if I'm correct on this or not.
if not, i'll try it again

#8
I thought there will be as ...

for vertical velocity, wr sinθ = v
but what now ?
if i consider pure rolling so
v sinθ = v
θ = 90
but that particle has zero velocity not normal ... :(

 

[SammyS]

#1
I guess no. coz considering anybody from a point inside or not it ... its in pure rotation and for a point outside it ... it wil have both translational and rotational motion
I agree.
#2
i don't know what they are asking .. if question is that can we write
A/a = v/w as (A+a) / (A-a) = (v+w) / (v-w) they yes. but won't that be a maths question
Think about the units of the things that are added together.
#3
Don't know ... some help please
I'd have to work out some details.
I do know this: It's more dangerous if the ladder slips when you're near the top!

#4
Don't know this either. yes i think that net torque will be zero but don't know how the it becomes horizontal
I would have to see a sketch of the balance.
#5
My attempt:
I (moment of inertia) along BY is less .. so kinetic energy along BY is less and thus its easier to rotate along BY
Yes. The moment of inertia about point B is less than that about point A.

#6
If ice melts ... radius increase ... as angular momentum is constant, I inc. and thus w dec ... so length of day-night inc
right? Correct!

#7
I guess solid sphere ... not just guess but worked out something ... someone just tell me if I'm correct on this or not.
if not, I'll try it again
Correct again. For a given radius and a given mass, the moment of inertia is least for the sphere.
If you do a detailed calculation of the acceleration the C.M. of each, both the radius and mass cancel out - much like mass cancels out for acceleration due to gravity.
#8
I thought there will be as ...

for vertical velocity, wr sinθ = v
but what now ?
if i consider pure rolling so
v sinθ = v
θ = 90
but that particle has zero velocity not normal ... :(
Assuming the frame of reference is the surface, then you are correct again. The instantaneous acceleration of the point of contact is vertical. As any point on the sphere approaches the surface, its velocity approaches the vertical, but (as I think you noticed) as that point comes in contact with the surface, its velocity becomes zero. A zero vector has indeterminate direction.

On the other hand: If the frame of reference has the same velocity as the C.M. of the sphere, then any point at the same height as the center of the sphere (except along the axis of rotation) is moving vertically for an instant.

See comments above in red.

 

[cupid.callin]

Thanks A lot for your reply Sammy

I really thought that after seeing so many questions, no one will reply ...

Thanks a lot :)

#2
Think about the units of the things that are added together.
i didn't noticed that
now i think its wrong ... we can't write it like that.

#3
I do know this: It's more dangerous if the ladder slips when you're near the top!
I guessed same coz it has happened with me
but i can't figure out why.

#4
I would have to see a sketch of the balance.
here's the sketch:

the horizontal bar holding 2 pans is hinged along a horizontal axis (perpendicular to the screen) ... so that it can rotate in vertical plane#8
Assuming the frame of reference is the surface, then you are correct again. The instantaneous acceleration of the point of contact is vertical. As any point on the sphere approaches the surface, its velocity approaches the vertical, but (as I think you noticed) as that point comes in contact with the surface, its velocity becomes zero. A zero vector has indeterminate direction.

On the other hand: If the frame of reference has the same velocity as the C.M. of the sphere, then any point at the same height as the center of the sphere (except along the axis of rotation) is moving vertically for an instant.


But will it be okay to take frame as CM?

and one more question,
what about rolling with slipping?
will velocity be vertical in that case?