- #1
geoffrey159
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Homework Statement
A block of mass ##m## slides on a frictionless table. It is constrained to move inside a ring of radius ##l## which is fixed on the table. At ##t=0##, the block is moving along the inside of the ring with tangential velocity ##v_0##. The coefficient of friction between the ring and the block is##\mu##. Find the velocity and position of the block at later times.
Homework Equations
Newton's second law of motion.
The Attempt at a Solution
Since I'm not sure what is meant by later times, I consider a small interval of time where the block sticks to the ring, say ##I=[0,T]##, and the block is under circular motion in that interval.
Since we want to know the speed which will have only tangential component in circular motion, we have to calculate ##\dot\theta##.
As initial conditions, I get ##\dot\theta(0) = \frac{v(0)}{r(0)} = \frac{v_0}{l}## and ##\theta(0) = 0##
Since the block is under ring reaction force radially, and under friction force tangentially, I get by Newton's second law that ##a_{\theta} = \mu a_r##, only for time ##t## in ##I##. Therefore I get the differential equation
## \frac{\ddot\theta}{\dot\theta} = -\mu \dot\theta ##
which thanks to the initial condition gives
## \dot\theta =\frac{v_0}{l} e^{-\mu\theta}##
That equation is equivalent to
##\frac{d}{dt} (e^{\mu\theta}) = \mu\frac{v_0}{l}##
Solving it gives
## \theta(t) = \frac{1}{\mu} ln(1+\frac{\mu v_0}{l}t)##
So that
## \dot\theta(t) = \frac{v_0}{l}\frac{1}{1+\frac{\mu v_0}{l}t}##
Therefore
## v(t) = r(t) \dot\theta(t) =\frac{v_0}{1+\frac{\mu v_0}{l}t}##
and
## \vec r(t) = l (cos(\theta) \vec i + sin(\theta) \vec j ) ##
3. Question
I begin in physics and I'm unsure about the solution, the question is not very clear either and I find the proof more complicated than other exercises in the book. Does it look good to you?
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