Chapter 5 - Circular Motion Flashcards
A 9.5-kg monkey is hanging by one arm from a branch and is swinging on a vertical circle. As an approximation, assume a radial distance of 85 cm between the branch and the point where the monkey’s mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 2.8 m/s. Find (b) the magnitude of the tension in the monkey’s arm.
For tension problems,
Fc = T - mg
and Fc = mv^2/r
Thus
T = mg + mv^2/r
T = 9.5*9.8 + 88N(from prev soln)
Uniform CIrcular Motion
Object travelling at constant (uniform) speed on circular path
Centripetal Acceleration:
Always points towards center and continually changes direction
Centriptal Force:
Force required to keep object moving at a speed on circular path
Fc = mv^2/r
Banked Curves & related equation
**Know this becuase not on formula sheet
tanx = v^2/rg where v is speed not velocity
Circular turn without static friction reqired to provide centripetal force, provided curve is banked at angle x
Satellites in Circular Orbit
Only one speed a satellite can have to remain in orbit with fixed radius
v = sqrt (GMe/r)
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a long wet patch in the road decreases the maximum static frictional force to one-third of its dry-road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
Centripetal force is caused by friction so you can set the following equations equal to each other
Fsdrymg = mv^2/r
Wet Road:
1/3Fsdrymg = mv’^2/r
You can cross out the mass on both sides because same. Plug in number 1 into number to get
1/3 (v^2/r) = v^2/r
Cancel out r and solve for v’
should get 12.12
On a banked race track, the smallest circular path on which cars can move has a radius of 112 m, while the largest has a radius of 165 m, as the drawing illustrates. The height of the outer wall is 18 m. Find (a) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.
First we need to find the banked angle.
The base of the triangle is 165-112 and the height 18m. The banking angle is tan-1 (18/53)
Thus use the equation tanx = v^2/rg to find v!
What provides the centrieptal force of an object is being supported by two strings?
Both the tension AND the gravity!
Outline how you would solve the following problem:
The 4.00 kg block in the figure is attached to a vertical rod 2m by means of
two strings (each 1.25m) When the system rotates about the axis of the rod, the strings
are extended as shown in Figure and the tension in the upper string is
80.0N. Tension in lower string and speed of the lock
First solve for the vertical forces
T1= tension upper
T2 = tension lower
T1cosx - T2cosx - mg = 0
mg = 39.24
You can solve for the angle using simple trig and get 36.87 (between vertical and Tupper). Sub in all the values and you can solve for T2 which is 30.95N
Now for the horizontal components, which provide centrieptal force
mv^2/r = T1sinx + T2sinx
You have all the values (can solve for r using trig and isolate for v) you should get 3.53 ish!
Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked
curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses
tires for which the coefficient of static friction is 0.85 on the same curve. What is the
maximum speed at which car B can negotiate the curve?
F = mac
which is also
F = mv^2/r
The force itself is the static friction force which is usFn
So
usFn = mv^2/r
. The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way
down the slope and is going around at a constant speed on a circle (r = 11.0 m) as the
carousel turns. The coefficient of static friction between the suitcase and the carousel is
0.760, and the angle in the drawing is 36.0°. How much time is required for your suitcase to
go around once?
So you want it set it up as so
mv^2/r us(mgcostheta)-mgsintheta
so you know how to get the rest which is pretty easy, the only confusing part is the centrieptal force
So it’s teh usFn but the normal is the horizontal component that keeps it from sliding down, teh static friction. and then you have to subtract the force of gravity as well which acts downward and is mgsintheta.
Ayou should solve to get a t of 45 seconds
Useful Insights for Banked Curve Problems
1) How to draw forces?
2) Min and max speed
- If force is parallel to the incline (like friction), the horizontal is cos
- If force is perpendicular to the incline (like normal) the horizontal is sine
- At minimum speed, the friction acts upward to prevent the car from sliding down
- At maximum speed, the friction acts downward to prevent the car from sliding upwards
Consider a circular curve of radius R = 200 m and bank angle β, where the
coefficient of static friction between tires and pavement is µs. A car is driven around the curve.
For a bank angle of β = 10° find the ratio of the maximum car speed vmax that puts the car on the verge of sliding out
for µs = 0.60 (dry pavement) to that for µs = 0.050 (wet or icy pavement)
This is a forces question, draw a free body diagram and consider the forces acting in the x and y direction.
In the x direction you got Nsintheta+usNcostheta = mv^2r
(aka the normal+friction) is the centrieptal force
In the y direction you got
Ncostheta-usNsintheta-mg = 0
(aka normal - friction - weight) and there’s no vertical acceleration
Now you want to factor out N for both and sub one into the other, isolate and solve and input both us values to find the ratio which should turn out to be
41.2/21.1 or 1.95!!!
A “swing” ride at a carnival consists of chairs that are swung in
a circle by 3.0 m cables attached to a vertical rotating pole, as the
drawing shows. Suppose the total mass of a chair and its occupant is
50.0 kg.
Determine the tension T in the cable attached to the chair
when its velocity is 3.46 m/s.
Hardest problem in the bokok
Vertical Force Equilibrium (no vertical acceleration so equals 0)
Tcosx - mg = 0
Tcosx = mg
T = mg/cosx (1)
Horizontal Force Equilibrium
Tsinx = mv^2/r
r = Lsinx
Tsinx = mv^2/3sinx (2)
Now sub 1 into 2
mgsinx/cosx = mv^2/3sinx
multiply both sides by cosxsinx
gsinx/cosx = v^2/3sinx
gsin^2x = v^2cosx/3
Sub in known values, and subsittut in sin^2x = 1-cos^2x
9.81cos^x + 3.99cosx - 9.81
cosx = 0.816
cosx = 0.816
solve for x which is 35.3
Sub back into equation 1 to find the tension which is around 600N
