- A wheel (mass M, radius R) rolls with speed v and angular velocity w (v=?R) and collides elastically with a wall. Immediately after the collision it has v away from the wall, but the same ? as before – so it’s skidding, not rolling.
Friction at the floor will eventually re-establish rolling motion, by doing negative work on the system and applying a torque that will change ?.
What is the new speed v’ away from the wall when rolling motion is re-established?
Answer by RickB
That’s an interesting question!
First, let’s set up direction conventions. Let’s say:
* The wall is on the left, so the wheel bounces off to the right. Let’s call “rightward” the positive linear direction.
* Let’s call “clockwise” the positive angular direction. (That means the wheel’s rotation is negative immediately after the bounce.)
During skidding, the force of friction is more or less constant (given by: (normal force)×(coef. of sliding friction)), so we can say that “Ff” is the (constant) force of friction during the skidding phase.
The force Ff produces a torque on the wheel, in the amount of: ? = (Ff)R. (This is a positive torque, since Ff points to the left and therefore tends to drive the wheel clockwise.) We know that the angular acceleration is given by:
? = ?/I = (Ff)R/I
where “I” is the wheel’s moment of inertia. I’m not sure what kind of wheel this is, but its moment of inertial can be represented as: I = ?mR², for some “?” that depends on how the wheel’s mass is distributed. (It’s some constant between 0 and 1.) So we can write:
? = (Ff)/(?mR)
At the same time, the ball’s linear acceleration is:
a = -Ff/m
(Here we use negative Ff, since the force points toward the left)
With these two values, we can write the ball’s angular and linear velocities as a function of time:
?(t) = ?0 + ?t = ?0 + (Ff)t/(?mR)
(where ?0 is the ball’s angular velocity right after collision)
v(t) = v0 + at = v0 + (?Ff/m)t
(where v0 is the ball’s linear velocity right after collision)
We are also given that ?0 = ?v0/R, so we can rewrite the ?(t) equation as:
?(t) = ?v0/R + (Ff)t/(?mR)
Now, at some time t?, rolling is re-established. That means:
v(t?) = ?(t?)R
v0 + (-Ff/m)t? = ?v0 + (Ff)t?/(?m)
t? = 2(v0)m?/[Ff(1+?)]
Now plug that value back into the “v(t)” equation:
v? = v(t?) = v0 + (-Ff/m)2(v0)m?/[Ff(1+?)]
= v0 – 2(v0)?/(1+?)
= v0(1 – 2?/(1+?))
So for example, if you assume that the wheel has a uniform density, then ?=½, in which case v? = v0/3.
- I own 1 unisex deep v and 1 unisex regular v-neck. I know that the regular v-neck runs slightly bigger than the deep, but how do they compare to the regular t-shirts as far as fit? I wear an xs in both v-necks for reference.
Answer by Benjamin
The fit of most of the crew necks (2001, BB401, TR401, 6401) is closer to the regular v neck (2456)
Go with XS in unisex styles and a S in womens.
- What current would flow in an ordinary 110 V household circuit, if a 1210 W hair drier and a 700 W microwave oven were operating simultaneously on this line?
Answer – 17.36 A
How much current would the hair drier and microwave from the previous question draw, if they were connected to the 110 V line in series?
Answer by Kim
IF you assume they act like simple resistors, then you can figure out their individual resistances using P = I^2 R. In the first part, you know P and I.
Once you know their resistances, when you put them in series you add their resistances. So then you can find I from I = V/R.
But I’m not sure these devices will really act like pure resistors because they have motors in them. There could be an inductive component.
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