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As I stated an earlier reply, friction acts even during rolling, as no body is perfectly rigid.

The disk has two horizontal forces acting on it.
1) Tension towards right
2) Friction towards right.

In this case, both turn too balance. (try checking ur self.) So, net force acting on the disc is zero... which means, the disc velocity remains constant...
Here initial velocity is zero... hence the disc's centre of mass continues to stay at rest.

Weight of a body in free fall is "mg" itself.

The apparent weight will be zero.

Ideally speaking, rolling friction will be zero on any perfectly rigid body.
But no body is perfectly rigid. There will be deformation at the point of contact with ground. So, the lines of action of normal force and weight are different. this results in rolling friction.

For detailed discussion, refer to:
http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html

The answer is

=v/r.
But the argument is as follows:
Let linear velocity of the centre of the disc be 'v1' towards left. Let the angular velocity of the disc be

.
The disc is moving to right with speed v. so the string moves with speed v to left.So, the velocity of the top most point of the disc should be v --------------1
velocity of the top most point wrt centre is

r towards left. the centre is moving to left with v1.
So the speed of top most point is v1+

r ---------------------2
by 1 and 2, we get,
v1 +

r = v -------------------3
Similarly we can say that the speed of bottom most point of the disc is

r-v1 to right ---------------------4
For the disc to roll on the block, the bottom most point should have the same speed as th eblock ---------------5
so, by 4 and 5, we haev

r-v1 = v ---------------------6

By 3 and 6, we get,

=v/r.

Method 1:

For every differential mass "dm" there is another identical "dm" exactly radially opposite. The centre of mass of these two exist at the centre of sphere. Arguing this way, the center of mass of sphere is at its centre.

method 2:

We can use Spherical polar coordinates to sole this.
Refer to
http://surendranath.tripod.com/Applets/Math/Coordinates/SphCoordApplet.html
for a visual description of spherical polar coordinates. Its just a matter of a triple integral to end up with the centre of mass to be at centre of sphere. Buzz if u find it difficult to integrate.

As stated by kmmankad, the man has a tendency to topple.
So, lets neglect the height of the man (a lilliput). So, no toppling!!

Now the question asks if the man can pull the block. The max possible friction is
f= (mu)*(M+m)*g, where mu is coeff of friction, m and m are masses of block and man respectively. So, if the man is sufficiently strong and can apply this much force on the srting (the same force will be transmitted to block in the form of tension), then he can definitely pull the block!!

kmmankad is wrong. The expression he stated is correct only when the block is constrained to move vertically. but here that is not the case in this problem.

waterdemon is perfectly right but the value of 'a' we got finally is not the absolute acceleration of block. It is the acceleration of block with reference to wedge. So, all this rel acceleration vector to the wedge's acceleration vector to get block's acceleration vector. It is essential to add the accelerations vectorially, not algebraically.

THIS IS NOT A DUMB QUESTION.
Every student will face the same problem during exams.

If u are in an objective type exam, then u are help less. The only thing u can do is guess an answer after eliminating few options.

If u are in a subjective type exam, then just put down till where u could go. If ur approach is right, u ill be given at least partial marks. I don't know if IITJEE mains will be back again, but if it is, then the following will prove to be very useful:
More weightage will be given to ur approach tan ur working. Will u believe, if I say that IITJEE-2001 toppers didnt solve even a single problem till the end??? He just gave the approach and left calculations, using the time remaining to think of approach than just completing calculations mechanically. So, approch is much more important than completion. if ur approach is right but got struck some where, u will definitely be given marks.

For problem solving hints, visit

http://www.oberlin.edu/physics/dstyer/SolvingProblems.html

if u want to be very clear in concept, then go for "Resnink and haliday" or "Young and Freedman". These are the best books for theory.

HC Verma may not be use fult for u as of now, as u said u are not clear abt the concept.

if u want to learn just waht is required for problem solving, then go for DC Panday.

If I say, 15 hours of mugging is necessary to get into IIT, are u going to get so much time into ur ife every day????

So, stop thinking of such questions. Work as much as u can. Always try to make best use of time.
http://www.goiit.com/posts/list/counselling-i-am-in-trouble-pls-help-me-34547.htm#174191

Regardign book for maths, the best book i have ever seen for IITJEE maths is "Mathematics for IITJEE by RD Sharma". TMH and das gupta are also considered to be good. As u ar ein 12th right now, i suggest Das Gupta.

I sincerely suggest u not to follow solutions. that will take u no where. Try to crack problems on ur own --- If u are unable to solve even after many serious attempts, then discuss with ur friends --- if even that doesnt work, then ask ur teacher -- only if all the previous steps fail, u have to go to solutions as the initials steps makes u strong in problem cracking skill, though it takes a lot of time...

As far as the present pattern of JEE is concerned, HC Verma should be more than sufficient.
But if u want ur self to be very strong in concepts (which will help u in cracking advanced exams like olympiads), then "Physics by Young and Freedmann" and "Fundamentals of physics by Resnick and Haliday" are very good. I personally suggest the former one.
If u want to practice too many problems, the DC Pandey is the one.
If u are in deep love with physics and want to get away from it, then catch hold of Irodov, which has toughest of problems.