in this problem we're studying a heavy disc that's accelerated from Rest by a string wrapped around a
small Hub that's glued on the center and we're given the specs on the disc it's three and a
half kilograms it has a radius of 40 centimeters and we're told that we apply a constant force by
pulling on that string so that's a constant force of 25 Newtons we're also told how much string we
pull out during the process that's 1.2 meters and how long it takes that's 1.5 seconds so in part A
we're going to find the work done as we pull that string out in Part B we'll use the work energy
theorem to get the final angular velocity of the disc in part C we compute the angular acceleration
of this disk and then Part D is a really unusual question we actually have everything we need
to compute the radius of that small Hub at the center so the work calculation is simple enough
because our force is pulling exactly parallel to the displacement as we pull this string out so
our work is just going to be that parallel Force multiplied by the displacement so F parallel times
Delta X the force was 25 Newtons the displacement 1 1.2 meters and when we run the numbers on this
we get 30 newton meters or joules in Part B we want to get the final angular velocity of the disc
and this is just a work energy theorem problem the work that we put into it is reflected in the
rotational kinetic energy of the disk there was no initial rotational kinetic energy so it's just
the work that we put in is equal to the final kinetic energy and that's given by one half I
Omega squared now we are reminded in this problem that the moment of inertia of a disk is one half
m r squared so we sub that in and I'll go ahead and just pop numbers in right now and then we'll
solve for Omega so the work that we put in was 30 joules then I have a one-half times one half
or one-fourth the mass of the disc 3.5 kilograms the radius was 0.4 meters just switching to SI
units there and we have to square that and then we have an Omega squared so we divide
30 by the product of all these coefficients and then square root the result to solve for
Omega and when I run the numbers on this I get 14.6 radians per second next we want to get the
angular acceleration of the disc and there's a subtle point that I want to highlight here the
force is constant that we're applying on this Hub that means the torque we're applying is constant
which in turn means the angular acceleration is constant so to get the angular acceleration at any
moment in time all we have to do is look at the simple rate of change in the angular velocity so
Delta Omega divided by delta T well the change in angular velocity is just the final minus initial
but the initial was zero so I have 14.6 radians per second that's divided by how long it took
to get up to that speed and that was given in the problem that's 1.5 seconds when I run the numbers
on this Alpha comes out to 9.73 radians per second squared finally we get to the somewhat unusual
Part D where we're asked to compute the radius of the Hub this is something we normally would be
given but in this problem we're actually finding it so we have to think about why this would be
possible and it starts with the rotational equivalent of Newton's second law so the net
torque is equal to I Alpha the moment of inertia times the angular acceleration well we're able to
compute the moment of inertia we already computed the angular acceleration and then I think about
well how can I break that torque down into smaller pieces that results from a force exerted through a
lever arm where the lever arm is actually the radius of the Hub that the string is wrapped
around so here's the key to the problem the torque we're exerting is given by the known force of 25
Newtons multiplied by the lever arm the radius of the Hub our moment of inertia is one half Mr
squared where R is the radius of the disk and then Alpha is something we already computed so now we
can just go ahead and plug in the numbers and solve for that little r the radius of the Hub F
was 25 Newtons so I have 25 R is equal to one half times the mass of the disc that was 3.5 kilograms
times its radius squared 0.4 squared times the angular acceleration Alpha which we computed was
9.73 radians per second squared and when I run the numbers on the this I get the r is 0.109 meters
which might be nicer to express in centimeters as 10.9 centimeters and we're done
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