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Getting Angular Speed Problem Wrong, Not Sure Why?
A small block on a frictionless, horizontal surface has a mass of 2.50×10−2kg . It is attached to a massless cord passing through a hole in the surface (the figure (Figure 1) ). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 1.85rad/s . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
What is the new angular speed?
Here are my steps (note: W is used for omega).
For the larger radius:
W=v/r
v=Wr
v=(1.85)(0.3)
v=0.555m/s
Now substitute v into the smaller radius:
W=v/r
W=(0.555)/(0.15)
W=3.7rad/second
This is exactly half of the correct answer, which is 7.4rad/sec. I have NO IDEA why this is happening, and I'm getting really frustrated. Any help is appreciated.
3 Answers
- Dr. ZorroLv 78 years agoFavorite Answer
The physics of it is that because the applied force has no torque relative to the center of rotation, the angular momentum must be constant. Angular momentum L is
L = I ω,
where I the moment of inertia and ω the angular velocity.
The moment of inertia for the point mass m at distance r moving in a circle is
I = m r^2
and so
L = m r^2 ω
Because L is constant, when we halve the radius, the angular velocity must become four times as large. So the angular velocity will be
ω = 4 * 1.85 rad/s = 7.40 rad/s.
The mistake you made is that you keep the velocity of the mass constant. That's not correct, the velocity will double.
- oldprofLv 78 years ago
We invoke the conservation of angular momentum. L = Iw = iW = L; I = kmR^2 and i = kmr^2.
So W = (I/i)w = (R/r)^2 w = 2^2 * 1.85 = 4*1.85 = 7.4 rad/s. ANS.
You failed to see the physics...the conservation of angular momentum. When you start a physics problem, before you put pencil to paper, check yourself by asking, what's the physics here?
- DambarudharLv 78 years ago
Where is your figure 1 ?
Nothing can be said without examining the figure.
