Tangential Acceleration of a Pendulum

[Sandman327]

Homework Statement

A 2kg pendulum bob on a string 2 meters long is released with a velocity of 1.5 meters/second when the support string makes an angle of 30 degrees with the vertical. What is the tangential acceleration of the bob at the highest point of its motion? The answer was given as 5.77 meters/second squared so I know that that is the answer but it's getting to that number that is giving me tons of trouble.

Homework Equations

Tangential acceleration = r(delta omega/change in time)

The Attempt at a Solution

I have an FBD and the only force acting on the bob is gravity. However the problem I'm having is that I am almost positive there is another equation I should be using but I'm not sure what it is or what I should try to solve for. Time? Maximum height? both?

 

[LowlyPion]

Consider the total energy of the system.

They give you the kinetic energy. You have the potential energy from gravity with reference to how high it is from the bottom. So figure the maximum height first, and then figure what angle that makes, and that angle to gravity is the fraction of gravity which is the tangential acceleration I think they are looking for.

 

[nlightner]

The tangential acceleration of a pendulum is given by the equation a = rω^2, where r is the length of the string and ω is the angular velocity. In this case, the angular velocity is changing as the pendulum swings, so we can use the equation a = r(dω/dt) to find the tangential acceleration.

To find dω/dt, we can use the conservation of energy equation for a pendulum: E = 1/2 m v^2 + mgh, where m is the mass, v is the velocity, and h is the height. At the highest point of the pendulum's motion, the velocity is zero, so we can set E = mgh. We also know that at this point, the height is equal to the length of the string, so h = r. This gives us the equation E = mgr.

We can also use the equation for the total energy of a pendulum, E = 1/2 m v^2 + 1/2 Iω^2, where I is the moment of inertia. For a simple pendulum, the moment of inertia is equal to mr^2. Substituting this into the equation and setting it equal to mgr, we get 1/2 m v^2 + 1/2 mr^2 ω^2 = mgr. Solving for ω, we get ω = √(2g/r).

Finally, we can substitute this value for ω into the equation for tangential acceleration, a = r(dω/dt), giving us a = r(√(2g/r))(d/dt)(√(2g/r)). Simplifying, we get a = 2g. Plugging in the given values of r and g, we get a = 2(9.8) = 19.6 m/s^2. This is the tangential acceleration at the highest point of the pendulum's motion.

It is important to note that the tangential acceleration is constant throughout the pendulum's motion, as long as the length of the string and the acceleration due to gravity remain constant. This means that the tangential acceleration will be the same at any point in the pendulum's swing, including the highest point.