Calculating velocity/acceleration from sway of an object.

[Gheret]

Homework Statement

A train accelerates from rest causing the handles that hang from the roof to sway 32 degrees from vertical. How long will it take the train to travel 300metres.

Homework Equations

The Attempt at a Solution

All I know is that the horizontal component of the force vector on the handle would be used to find acceleration.

 

[gneill]

Did you draw a Free Body Diagram for a typical handle? What is the relationship between the acceleration due to gravity and the horizontal acceleration of the train in terms of the hang angle?

 

[Gheret]

Thanks for the reply.

Would it be tan32=?/g
With ? being the acceleration of the x component.

 

[gneill]

That sounds reasonable.

 

[rwisz]

I would approach this problem by first understanding the concept of velocity and acceleration. Velocity is the rate of change of an object's position over time, while acceleration is the rate of change of an object's velocity over time. In this scenario, the train is accelerating from rest, which means it is increasing its velocity over time.

To calculate the velocity and acceleration of the train, we need to use the equations of motion. The first equation, v = u + at, relates velocity (v) to initial velocity (u), acceleration (a), and time (t). Since the train starts from rest, its initial velocity (u) is 0. Therefore, the equation becomes v = at.

The second equation, s = ut + 1/2 at^2, relates displacement (s) to initial velocity (u), time (t), and acceleration (a). In this equation, we can substitute the value of velocity (v) from the first equation, which gives us s = 1/2 at^2.

Now, we can use the given information to solve for time (t). The train travels 300 meters, which is equal to the displacement (s). The angle of sway, 32 degrees, can be used to find the horizontal component of the force vector, which is equal to acceleration (a). Using basic trigonometry, we can calculate that the horizontal component of the force vector is approximately 0.55 times the total force.

Substituting these values into the second equation, we get 300 = 1/2 (0.55a)t^2. Simplifying, we get t^2 = (600/a). To solve for time (t), we need to find the value of acceleration (a). We can do this by using the given angle of sway and the formula for centripetal acceleration, a = v^2 / r, where r is the radius of the train's circular motion. Since the train is moving in a straight line, we can assume that the radius is equal to the length of the train car, which is approximately 20 meters. Therefore, a = (v^2 / 20).

Now, we can substitute this value of acceleration (a) into our equation and solve for time (t). We get t = √(600 / (v^2 / 20)). To calculate the value of velocity (v), we can use the first equation