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 = 11.2 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 37.0°. How much time is required for your suitcase to go around once?
To find the time required for the suitcase to go around once, we can use the centripetal acceleration formula:
a = (v^2) / r
where:
a is the centripetal acceleration,
v is the linear velocity, and
r is the radius of the circular path.
Given:
r = 11.2 m
The linear velocity can be found using the formula:
v = ω * r
where:
ω is the angular velocity.
To calculate the angular velocity, we can use the formula:
ω = θ / t
where:
θ is the angle covered in radians, and
t is the time taken to cover that angle.
Given:
θ = 37.0°
We need to convert it to radians by multiplying it by (π / 180).
θ = 37.0° * (π / 180) = 0.6458 rad
Substituting the given values, we have:
v = ω * r
ω = θ / t
a = (v^2) / r
We need to find the time (t), so let's solve for t:
ω = θ / t
t = θ / ω
Now, let's substitute the known values:
t = 0.6458 rad / ω
To find ω, we need the linear velocity (v). Let's calculate that:
v = ω * r
v = ω * 11.2
Now, let's find ω by substituting the linear velocity equation into the ω equation:
t = 0.6458 rad / (v / 11.2)
Now, let's substitute the equation for v:
t = 0.6458 rad / ((ω * 11.2) / 11.2)
t = 0.6458 rad / ω
Now, we can substitute the value of the coefficient of static friction to solve for ω:
t = 0.6458 rad / (0.760 * g / r)
where g is the acceleration due to gravity.
Finally, to find the time required for the suitcase to go around once, we need to calculate the value of ω and substitute it into the equation:
t = 0.6458 rad / (0.760 * 9.8 m/s^2 / 11.2 m)
Simplifying the equation, we have:
t = 0.6458 rad / 0.6522 rad/s
t ≈ 0.990 s
Therefore, it would take approximately 0.990 seconds for your suitcase to go around once.
To calculate the time required for the suitcase to go around once, we can use the formula for centripetal acceleration:
a = (v^2) / r
where "a" is the centripetal acceleration, "v" is the velocity, and "r" is the radius of the circle.
First, we need to find the centripetal acceleration. The centripetal acceleration is the net force acting towards the center of the circle divided by the mass of the object.
The net force can be found using the equation:
F_net = μ_s * N
where "F_net" is the net force, "μ_s" is the coefficient of static friction, and "N" is the normal force.
In this case, the normal force is equal to the weight of the suitcase since it is on a flat surface.
N = m * g
where "m" is the mass of the suitcase and "g" is the acceleration due to gravity (approximately 9.8 m/s^2).
Now we can substitute the values into the equation for the net force:
F_net = μ_s * m * g
Next, we can calculate the centripetal acceleration using Newton's second law:
F_net = m * a
Setting the equations for net force and centripetal acceleration equal to each other, we get:
μ_s * m * g = m * a
Simplifying, we have:
a = μ_s * g
Now, we can substitute this value into the centripetal acceleration equation:
μ_s * g = (v^2) / r
Rearranging the equation to solve for velocity, we get:
v = sqrt(μ_s * g * r)
Finally, we can use the formula for time:
t = (2 * π * r) / v
Substituting the values we know:
t = (2 * π * 11.2) / sqrt(0.760 * 9.8 * 11.2)
Calculating this value will give us the time required for the suitcase to go around once.
21.6 seconds
The trick t this problem is to realize that the suitcase iss ON THE VERGE OF slipping.
The static coefficient of friction detremines the maximum possible friction force wthout slipping. The actual force can be much less.
Based upon the observation that the suitcase slipped part way down but not all the way down the sloped carousel, you can infer that the friction force is the maximum value, which in this case is M g cos 37 * mus. The net inward horizontal force equals the centripetal force. Both friction and gravity contribute to the inward force.
M g cos^2 37 mus + M g sin 37 = M V^2/R
"mus" is the static coefficient of friction
I had to square the cosine term because the friction force does not act directly inward.
Cancel the M's, and solve for V. Then get the time T for one trip around the carousel from V T 2 pi R