At the point 𝐴 the surface of the cylinder is stationary because it's in contact with the stationary ramp. So we can take this as the centre of rotation at the instant shown.
If we calculate the torque about A then we can use: τ = Iₐα to calculate the angular acceleration about A, where Iₐ is the moment of inertia about A. We can can use the parallel axis theorem to find Iₐ
In this case we are considering the rotation of the cylinder about the point A, so Iₐ is the MOI of the cylinder about a point on its circumference: Iₐ = ¹⁄₂mR² + mR² Yes?
The value of μₛ we get this way is the minimum possible value of μₛ i.e. μₛ could be higher but it couldn't be smaller or the cylinder would slide instead of rolling.
Point (b) : $fR = I \alpha$ Where $\alpha = \frac{a}{R}$ $fR = \frac{mR^2}{2} \cdot \frac{a}{R}$ $f = \frac{ma}{2}$ Now I write the force diagram On the x axis : $-f + mg\sin(\theta) = ma$ On the y axis: $N = mg cos(\theta)$ So : $-\frac{ma}{2} + mgsin(\theta) = ma$ $a = \frac{2}{3}g \sin(\theta)$ $f = f_s = \mu_s \cdot N$ But : $f = \frac{ma}{2}$ $\mu_s mg \cos(\theta) = \frac{ma}{2}$ $\mu_s = \frac{ma}{2(mg\cos(\theta))}$
In the last three lines that $a$ I write would be the $a$ that I calculated above
@JohnRennie However, I read that below thsi exerercise It gave me this suggestion:
Tip: the moment of inertia of a cylinder with respect to the axis coinciding with the axis of longitudinal symmetry passing through the center of mass (the one perpendicular to the bases) is equal to $ I = \frac{mR^2}{2}$
Ah nothing, I thought It was referring to point c
I'll let you know if I find anything on this point
So if the frictional force was non-zero that would mean there must be a non-zero horizontal force so the cylinder must be accelerating or decelerating. And that must mean its KE is changing. And that must mean something is doing work on it. Yes?
But in pure rolling the bottom of the cylinder is stationary because it is in contact with the stationary floor. And that means any force acting at the contact point cannot do any work. Work can only be done by friction when the two surfaces slide over each other.
But if no work can be done that means the KE of the cylinder cannot change i.e. it must be moving at constant velocity.
And that means the net horizontal force must be zero.
@JohnRennie Please have a look at this : https://datascience.stackexchange.com/questions/129754/daily-balance-prediction-using-lstm-arima?noredirect=1#comment128372_129754 Any ideas on how to increase the prediction accuracy?
Problem Solving Strategies
Problem Solving Strategies