Question

A block of mass 1.36kg is on a wedge with a block-wedge static coefficient of friction, μs = 0.214. The wedge angle θ = 31.5o is as shown in the above illustration. The wedge is rotated with an angular frequency of ω with the block's center of mass located a distance L = 0.672m up the ramp from the axis of rotation as depicted by the above diagram. (a) What is the largest positive value of ω that is allowed to keep the block at the same location on the wedge? (b) What is the smallest positive value of ω that is allowed to keep the block at the same location on the wedge?

Answer 1

My question is, how does the angular frequency come into this?

Answer 2

hmm maybe with torque ?

Answer 3

Believe it or not, this does not involve torque.

Answer 4

ω its angular velocity, isnt it? and ωr = v. and fnet in a circular movement is F=m*a, where a is v^2/r

Answer 5

yes

Answer 6

if you do newton's second law equations for x, it gives that Fnet= -Fgx + Ffriction.
thus
\[m \times v ^{2}/r = -mgsin \theta + \mu mgcos \theta\]
mass cancels, isolate v and use ωr = v to find ω

Answer 7

this is if the ω is at its lowest, because the friction opposes the motion that would occur if there was no friction. thus, its pointed upwards. this equation would give you the lowest ω possible. if you want the highest possible, put a " - " in front of the friction force, because it would slide upwards and the friction would be negative

Answer 8

where did you get that the net force = -force of gravity (wrt x) + force of friction?

Answer 9

net force is sum of the forces, just put the right signs in front

Answer 10

and Fgx is the gravity component on x axis, and x axis is along the slope of the block

Answer 11

If you solve for v, you get a negative under the radical.

Answer 12

what r did you use?

Answer 13

0.672

Answer 14

i think its L * cos(theta)

Answer 15

oh ya its -Fnet, because fnet points towards the center of the circle and its at the left so its negative sorry

Answer 16

and r = L, correct?

Answer 17

There are two mistakes here.

1) From, $$F=\mu_sR$$and $$R\not = mg\cos\theta$$ (since there is an acceleration on y direction)
therefore ,
$$F\not =\mu_smg\cos\theta$$
2) If you apply F=ma along x axis, you should consider the acceleration component towards that axis. Therefore,
$$a_x\not = \frac{v^2}r$$

So what I would do is,
apply F=ma vertically,
$$\begin {align*}
R\cos\theta&=F\sin\theta+mg\\
R(\cos\theta-\mu\sin\theta)&=mg\\
R&=\frac{mg}{cos\theta-\mu\sin\theta}
\end {align*}$$

applying F=ma horizontally
$$\begin{align*}
F\cos\theta+R\sin \theta=mr\omega\\
R(\mu\cos\theta+\sin\theta)=mr\omega \end{align*}$$
substitute R from previous equation and find omega