physics related numericals

Question

a block of mass m is kept on horizontal turn table at distance r from centre of the table. coeff of friction b/w table & block is μ.now turn table starts rotating with angular acceleration alpha. find the time after which slipping occur b/w table & block.
so in the above question, a(net) is used. why??
explain the method & if u use a(net) then tell the reason why a(net) is used?

Answer 1

Hi Prashant!

 

An object moving on a circular path must have one acceleration on it -- the centripetal acceleration -- but may also have a second acceleration, always pointing in its direction of motion, if it is changing speed while on its circular path.  That is the case in this problem, since the block is speeding up.  These two accelerations are, at every point, perpendicular to each other. The centripetal acceleration points toward the center of the circle, and so is in the "radial" direction.  The acceleration that changes its speed points along a line in its direction of motion at all times, a line that is always tangent to the circle, and is thus in the "tangential" direction

 

Because these two accelerations are always perpendicular (the tangential and radial directions in circular motion are always perpendicular), we can combine them -- as components -- into what your problem (probably) calls a_net.  This is the net acceleration the block has to have, and thus determines the net force that has to be provided.  If static friction is providing the force, the maximum a_net that can be sustained is determined by the maximum friction force.

 

We can label the centripetal acceleration (in the radial direction) a_c and the tangential acceleration a_t.  Then, we can write expressions for each based on standard definitions:

 

a_c = rω²

 

where r is a given quantity and ω = angular velocity at a some instant

 

a_t = rα

 

where r and α are both given quantities

 

To determine the magnitude of the net acceleration, we have to take these two perpendicular accelerations as components and combine them into a resultant acceleration using the Pythagorean theorem:

 

a_net = √((rα)²+(rω²)²)

 

Hence, the net force that must be on the block at every instant to keep it moving in this way is:

 

F_net = ma_net = m(√[(rα)²+(rω²)²])

 

Note that this force increases with increasing angular velocity.

 

This force must be provided by (static) friction.  For (static) friction coefficient μ, the maximum possible static friction force is given by

 

F_fmax = μN

 

where N = normal force on the block.  Since the block is just sitting on a horizontal surface, the normal force must equal the weight, mg.  Thus:

 

F_fmax = μmg

 

Once the net force required to keep the object speeding up on its circular path exceeds F_fmax, the block must start to slide, as (static) friction can no longer hold it on its path.  Now we can solve for the value of ω that makes this happen, when:

 

F_net = F_fmax

 

m(√[(rα)²+(rω_max²)²]) = μmg

 

The first thing to note is that the mass of the block cancels out, and thus does not matter in this context.  Any mass with the same coefficient of friction would be affected the same way.

 

You just have to invert this equation to solve for ω_max in terms of a bunch of other quantities that are given or known.  When I did this, I obtained the unwieldy but manageable expression:

 

ω_max = [(√[(μg)²-(rα)²])/r]^1/2

 

Now we just have come up with an expression for the time at which the turntable reaches this angular velocity.  For this, we can use rotational kinematics (the rotational analog to one-dimensional linear kinematics).  In the group of rotational kinematics equations, there is one that states:

 

ω = ω_o + αt

 

where ω_o = initial angular velocity

 

Because we are told the turntable "now...starts rotating," we can take it that the turntable started from rest, meaning ω_o = 0.  Therefore, the kinematic equation becomes:

 

ω = αt

 

meaning

 

t = ω/α

 

And the time at which the block starts to slip is then

 

t_slip = ω_max/α

 

Just put in the expression for ω_max from above, and that should complete the problem.  

 

I hope this helps!  Let me know if you have any questions or problems with any part of this.

Answer 2

The block will slip when the force due to the rotation equals the frictional force that the block produces.

 

Force due to rotation: Frot = ma. Here, a is given by a  = v^2/r.

 

Acceleration: The acceleration increases with time, since we are given its angular acceleration α. So the instantaneous angular velocity is v = αt.

 

The block has a normal force of Fn = mg. The frictional force that the block can provide is therefore Ff = µFn = µmg.

 

Now it's just a matter of equating the force due to rotation Frot to the frictional force Ff. 

 

Frot = m*v^2 / r = m*(αt)^2 / r.

 

Equating:

 

Frot = Ff

m*(αt)^2 / r = µmg

 

Solve this for the time, t.