so Στ = Ffr * R = I*α
and ΣF = Fx - Ffr = macm
solving for acm = (Fx - Ffr) / m
and plugging this into the torque equation Ffr = βmacm = β(Fx - Ffr)
so Ffr = (βFx)/(1 + β)
?
Homework Statement
A block of mass m is on a platform of mass M, supported by a vertical, massless spring with spring constant k.
When the system is at rest, how much is the spring compressed?
When the spring is pushed an extra distance x, what is the frequency of vertical oscillation...
So aCM = -kx/m, and because x = A in this problem, then my answer for the wheel's center of mass acceleration is simply aCM = -kA/m
And because aCM = R*α, then the angular acceleration of the wheel α = aCM / R, so α = (-kA)/(mR)
And substituting this relationship into the Ffr = βmRα equation...
So Energy after the collision = (1/2)(2m)(v0^2/4) + (1/2)kA0^2
Substituting A0 = √(mV0^2)/k
Then, E = mv0^2/4 + mv0^2 = (1/2)kA^2
Simplifying: E = (mv0^2)/2 + 2mv0^2 = kA^2
So A^2 = (5mv0^2)/2
And because the coefficient of A0 is just one, the ratio* is (5/2)?
So because the blocks collide at maximum potential energy, can I relate that direction to maximum kinetic energy?
(1/2)kA0^2 = (1/2)mv0^2
So A0 = √(mV0^2)/k ?
Ahh, right, I forgot about the friction in torque.
So Στ = F(friction) = Iα
F(friction) = βmR^2 * α
But how do I connect that back to the Fx the question is asking for?
Homework Statement
A block os mass m is attached to a horizontal spring, which is attached to a wall. The block is oscillating without friction with initiation amplitude A0 and maximum velocity v0. When the block is at its maximum amplitude (and therefore instantaneously at rest), is it struck...