Answer Conical Pendulum: Determine Angle in Car Traveling Circular Road

[neoh147]

what is this topic??

A simple pendulum is suspended inside a car. The car then travels around a flat circular road of radius 350m at a constant speed of 30 ms^-1 Determine the angle which the pendulum string will make with the vertical.

this question is related to which topic?? (is it conical pendulum??)

please help me to answer this question...

 

[dacruick]

This is actually pretty simple. There is a constant acceleration towards the centre of the circle, so this question can be simplified to a static situation. The pendulum is not moving back and forth, it is suspended at a constant angle. This angle is related to the acceleration(which is the same acceleration you feel when a car is turning on a freeway ramp). Further, it is a simple pendulum so there is a point mass 'm' at the end of a weightless shaft of length 'd'. You have to figure out the equilibrium position of this pendulum, which is the angle at which the torque due to gravity on the pendulum is equal to the torque due to the circular motion.

Steps:
1)There are two forces, one is gravity, and the other is caused by circular motion. You know the value of gravity, so find the force caused by circular motion.
2) Find the torque on the pendulum due to gravity (in terms of the angle). To clarify this, if the pendulum is sitting vertically downwards, there will be no torque due to gravity because the angle with the vertical is 0. This is the equilibrium position if the car isn't accelerating.
3) Next you have to find the torque on the pendulum due to your circular motion.
*be careful here with your sines and cosines*. The force of centripetal acceleration is towards the centre of the circle (in the plane of the ground), while the force of gravity is downwards(making a 90 degree angle with the ground).
4) After you have found each of these equations, you can equate them by the condition that the torques are equal, and solve for the angle.

Above are good mathematical steps, but if this type of problem is something new to you, you should draw out a really nice Free Body Diagram. It will help you sort out the angles and understand the problem more thoroughly.

 

[neoh147]

just use horizontal component

$mgtanθ-ma=0$ [a=centripetal force]

and

$a=\frac{v^{2}}{r}$

to solve the eqn right??

$θ=tan^{-1}(\frac{a}{g})$
$θ=tan^{-1}(\frac{v^{2}}{rg})$
$θ=tan^{-1}(\frac{(30ms^{-1})^{2}}{(350m)(9.81ms^{-2})})$
$θ=14.7$

 

[dacruick]

I didn't plug in your numbers but your original equation is correct. I don't know how you came to the conclusion that you did, but its right.