A popular carnival ride consists of seats attached to a central disk through cables. The passengers travel in uniform circular motion. As shown in the figure, the radius of the central disk is R0 = 3.00 m, and the length of the cable is L = 1.43 m. The mass of one of the passengers (including the chair he is sitting on) is 56.1 kg.
a) If the angle θ that the cable makes with respect to the vertical is 30°, what is the speed, v, of this passenger?
b) What is the magnitude of the force exerted by the cable on the chair?
I got b), F=635.5 N, but I can't figure out a. I tried setting 635.5=mv^2/r, but that didn't work. I also tried mg=mv^2/r, v=sqrt(gr) but that also didn't work. Any help or direction would be greatly appreciated!
Here's the figure: i41.tinypic . com/2ho9v69.png
I cannot find the web site of your figure, but thanks for showing you work.
Let the cable force be T.
T cos30 = M g
T = 634.8 N
T sin30 = M V^2/R = T/2
R is the orbit radius, which is probably
R = Ro + Lsin30 = 3.72 m
(Without the figure, that is an educated guess)
Solve for V
Thank you so much, you guessed right without the figure. I really appreciate the help!
To find the speed of the passenger in uniform circular motion, you need to use the concepts of centripetal force and tension in the cable. Let's break down the problem step by step:
a) Finding the speed of the passenger:
By looking at the figure, we can see that the gravitational force mg acting on the passenger can be resolved into two components: one perpendicular to the circular path (mgcosθ) and the other parallel to the circular path (mgsinθ).
To keep the passenger moving in a circular path, there must be an inward force acting on the passenger known as the centripetal force. In this case, the tension in the cable provides the centripetal force.
The tension in the cable can be found using the component of the gravitational force acting in the radial direction (perpendicular to the circular path). This component is given by T = mgcosθ.
Now, equating the centripetal force (T) with the radial component of the gravitational force, we have:
T = mv²/R (Equation 1)
where T is the tension in the cable, m is the mass of the passenger, v is the speed of the passenger, and R is the radius of the circular path.
We know the length of the cable (L) and the radius of the central disk (R0). By using trigonometry, we find the radius of the path (R) as:
R = √(R0² - L²) (Equation 2)
Substituting Equation 2 into Equation 1, we get:
mgcosθ = m(v²/√(R0² - L²))
Simplifying and solving for v, we get:
v = √[(gR0² - gL²) / (R0² - L²)] (Equation 3)
Now, you can substitute the values given in the problem (R0 = 3.00 m, L = 1.43 m, θ = 30°, g = 9.8 m/s²) into Equation 3 to find the speed of the passenger (v).
b) Finding the magnitude of the force exerted by the cable on the chair:
To determine the magnitude of the force exerted by the cable on the chair, we need to find the tension (T) in the cable.
From Equation 1, we have:
T = mv²/R
By substituting the values of m, v, and R obtained from part a) into this equation, you can calculate the tension in the cable (T).
Hope this helps! Let me know if you need any further clarification.