- A wheel (mass M, radius R) rolls with speed v and angular velocity w (v=ωR) and collides elastically with a wall. Immediately after the collision it has v away from the wall, but the same ω 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 ω.
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: τ = (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: α = τ/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 = βmR², for some "β" that depends on how the wheel's mass is distributed. (It's some constant between 0 and 1.) So we can write: α = (Ff)/(βmR) 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: ω(t) = ω0 + αt = ω0 + (Ff)t/(βmR) (where ω0 is the ball's angular velocity right after collision) and: v(t) = v0 + at = v0 + (−Ff/m)t (where v0 is the ball's linear velocity right after collision) We are also given that ω0 = −v0/R, so we can rewrite the ω(t) equation as: ω(t) = −v0/R + (Ff)t/(βmR) Now, at some time t′, rolling is re-established. That means: v(t′) = ω(t′)R or: v0 + (-Ff/m)t′ = −v0 + (Ff)t′/(βm) From which: t′ = 2(v0)mβ/[Ff(1+β)] Now plug that value back into the "v(t)" equation: v′ = v(t′) = v0 + (-Ff/m)2(v0)mβ/[Ff(1+β)] = v0 - 2(v0)β/(1+β) = v0(1 - 2β/(1+β)) = v0(1-β)/(1+β) So for example, if you assume that the wheel has a uniform density, then β=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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