A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 21.0° with the horizontal. A 37.0 kg piece of luggage is placed on the carousel, 7.10 m from the axis of rotation. The travel bag goes around once in 44.0 s. (a) Calculate the force of static friction between the bag and the carousel. b) (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to a position 7.88 m from the axis of rotation. The bag is on the verge of slipping as it goes around once every 28.20 s. Calculate the coefficient of static friction between the bag and the carousel.
For a I am getting 43 and I don't think that is right and for b I am getting 12 and I don't think that is right either. Are your values any where close to mine?
To find the force of static friction between the bag and the carousel, we can use the formula:
fs = m * ω^2 * r
where
fs is the force of static friction,
m is the mass of the bag (37.0 kg),
ω is the angular velocity (ω = (2π / T), where T is the time for one revolution),
r is the distance of the bag from the axis of rotation (7.10 m).
Let's calculate fs:
a) Calculating the force of static friction:
T = 44.0 s (time for one revolution)
ω = (2π / T) = (2π / 44.0) rad/s
r = 7.10 m
fs = (37.0 kg) * ((2π / 44.0) rad/s)^2 * (7.10 m)
fs = 133.12 N
Therefore, the force of static friction between the bag and the carousel is 133.12 N.
b) To calculate the coefficient of static friction, we need to find the new angular velocity (ω) and the new radius (r) of the bag.
T = 28.20 s (time for one revolution)
ω = (2π / T) = (2π / 28.20) rad/s
r = 7.88 m
Using the formula for static friction (fs = m * ω^2 * r), we can rearrange it to solve for the coefficient of static friction (μs):
fs = m * ω^2 * r
μs = fs / (m * ω^2 * r)
Let's calculate μs:
μs = (133.12 N) / ((37.0 kg) * ((2π / 28.20) rad/s)^2 * (7.88 m))
μs = 0.549
Therefore, the coefficient of static friction between the bag and the carousel is approximately 0.549.
I erred. centripetal force is directed outward, not down the slope, so
a. gravity down the slope=mgSinTheta
centripetal force : m*v^2/r * cosTheta
friction=gravity down+centripetal
v=2PI r/period
b. same logic, change r, and period.
for coefficent, friction then equals mu*mgCosTheta, solve for mu.
a. gravity down the slope=mgSinTheta
centripetal force : m*v^2/r
friction=gravity down+centripetal
v=2PI r/period
b. same logic, change r, and period.
for coefficent, friction then equals mu*mgCosTheta, solve for mu.