Help with circular motion question pleeease?

Question

Here's the question:

A fixed solid sphere with a smooth surface has centre O and radius 0.8m. A particle P is given a horizontal velocity of 1.2ms-1 at the highest point on the sphere, and it moves on the surface of the sphere in part of a vertical circle of radius 0.8m.

Find the tangential component of the acceleration of P at the instant when OP makes an angle π/6 radians with the upward vertical. (You may assume that P is still in contact with the sphere.)



Now I know that you resolve tangentially to get



mgsin π/6 = ma => a=g/2



But why does this method give the wrong answer:



At top, v=1.2 so ω0=1.2/0.8=1.5 rads-1

At θ= π/6, v=1.88ms-1 (this is correct – it was found in a previous part of the question) so ω= 1.88/0.8= 2.35 rads-1



So using ω2= ω02 +2αθ => α=(ω2- ω02)/(2θ) = (2.352-1.52)/(π/3)= 3.13rads-2

So tangential acceleration= rα= 0.8x3.13= 2.51ms-2



But 2.51=/= 4.9 so why is the second method wrong? Is α not the same as d2θ/dt2? (if not, then what is it?)


Thanks in advance! 
PS sorry i don't know how to get the super/subscript right on this.

oldprof · Accepted Answer

There are a whole lot of givens needed that were not given.  So I need to make some assumptions.  One, the only motivating force on that particle is the force of gravity.  And that's w = W sin(theta) where theta is the angle between P and it's starting point P'; that is, POP'  So when atop the sphere and POP' = 0, there is nothing to push the particle.  But when it's off to one side and POP' = 90, the full weight of the particle W is pulling on it.  

And that, unfortunately, means the tangential force and consequent acceleration are non-linear; so unless we're ready to do some integration, we are not going to use averages here.  That alpha in 2αθ is an average alpha that is not warranted.  Remember all those kinematic equations are based on distance = average speed X time traveled and its that average speed and its average acceleration Vavg = Aavg * T you are lacking when the acceleration is non linear over time and/or distance.

When POP' = pi/6 = 30 deg, the tangential pull is w = W sin(30) = W/2 so the tangential acceleration is At = W/2M = Mg/2M = g/2.  As you found.

troutfisher · Answer

If you do the geometry correctly, the normal component of gravity at pi/6 from vertical is -gsin(pi/6), and the tangential component is -gsin(pi/3) = -.866g

Help with circular motion question pleeease?

Related Questions

Trending Questions