Hello, I need your help with a Physics problem from AP Physics B. Thank you for your help!
A .50 kg pendulum bob is suspended from a 1.0m long string anchored to a bar that is 2.0m above the floor. The bob is pulled back (lifted) until the string makes a 40 angle with the vertical. The pendulum is released and allowed to swing past the vertical The string string is then cut by a razor when the string makes a 20 degree angle with the vertical. How far will the pendulum travel horizontally before hitting the floor?
We need to use the conservation of energy to find the bob's velocity when it becomes a free-falling projectile. I'm going to give you significant hints, but I'm going to leave significant chunks of work for you to do too. (I have faith in Minnesota Physics students -- I grew up in Moorhead.) There will be 3 triangles that I suggest you draw to confirm what I say. When the bob is held at the starting position, consider the right triangle formed by the string, a horizontal line from the bob to a vertical down from the anchor, and a vertical line down from the anchor to that horizontal line. The length, h1, of the side adjacent to the 40 degree angle is
h1 = 1 m*cos40
So the gravitational potential energy of the bob at the starting point, with the bottom of the swing 1 m below the anchor being the reference point, is
GPE1 = m*g*h2
where m = 0.50 kg, g = 9.8 m/s^2, and h2 = 1 m - h1. You can work out the value of GPE1 in Joules.
That is the energy available to the system when the bob starts swinging. That will be entirely in the form of kinetic energy as the bob swings through the lowest point of the arc. When it has gone past such that the string makes a 20 degree angle with vertical and the string is about to be cut, it again has some gravitational potential energy. Another triangle is needed to analyze this.
GPE2 = m*g*h3
where h3 = 1 m - 1 m*cos20. GPE2 has a smaller value than GPE1. The difference, GPE1-GPE2, is the value of kinetic energy the bob has. You can get the velocity using the equation
GPE1-GPE2 = KE = (1/2)*m*v^2
Solve for v.
Now we need to know the angle that the velocity, at the time of cutting the string, makes with horizontal. It will be along a line tangential to the arc at the position it was when the string was cut. So that tangential line is perpendicular to the string. Now consider the triangle, slightly larger than the last one, formed by the vertical line down from the anchor, the string, and a line perpendicular to the string and intersecting the vertical down from the anchor. So the right angle of this triangle is at the corner where the bob is. The angle a bit more than 1 m below the anchor is 90-20=70 degrees. So the launch velocity has an angle with horizontal of 20 degrees above horizontal.
Now it's an exercise in projectiles. Use Vi as the variable name of the velocity you determined 2 paragraphs above. The horizontal and vertical components of that are given by
Vih = Vi*cos20
Viv = Vi*sin20
Let both of these have a positive value. Therefore we need to consider the acceleration due to gravity to have a negative value. Find the time to the floor using the kinematic formula
y = Vi*t + (1/2)*a*t^2
where y = the height above the floor, Vi = Viv, and a = -g = -9.8 m/s^2. The height above the floor, y, might be hard to see. Compare what you think with what I think.
y = 2 m - 1 m + h3
Solve for the time for the bob to hit the ground.
The distance it will travel horizontally in that amount of time is just Vih*t. But, that flight didn't start immediately below the anchor. Perhaps you should include that in your result.
I hope this helps,