A simple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2). i don't knw where to start plzz help me
yes
KEinitial+ PEinitial=KEbottom
So the idea is to use geometry to find the initial height of the bob.
Would h= 2.0(1-cos25) ? check my thinking
ya thts rt . so h will be 0.187m
To solve this problem, we can use the principles of conservation of mechanical energy. The mechanical energy of the pendulum consists of two parts: potential energy and kinetic energy.
The potential energy is given by the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
The kinetic energy is given by the equation KE = (1/2)mv^2, where m is the mass and v is the velocity.
At the highest point of the swing (release point), all of the energy is potential energy, and at the lowest point, all of the energy is kinetic energy. Since there is no mechanical energy lost due to friction or other forces, the total mechanical energy remains constant.
Given that the length of the pendulum is 2.0 m, we can calculate the height at the highest point using trigonometry. The height is equal to the length times the sine of the angle from the vertical, so h = 2.0 * sin(25°).
To find the speed at the bottom of the swing, we need to find the kinetic energy at that point. Since the total mechanical energy is conserved, we can set the potential energy at the highest point equal to the kinetic energy at the lowest point:
mgh = (1/2)mv^2.
Now, we can solve for v:
(1/2)mv^2 = mgh.
Dividing both sides by m:
(1/2)v^2 = gh.
Now, we can plug in the known values:
(1/2)v^2 = (9.8 m/s^2) * (2.0 m * sin(25°)).
Simplifying:
v^2 = 9.8 m/s^2 * 2.0 m * sin(25°).
Taking the square root of both sides to find v:
v = sqrt(9.8 m/s^2 * 2.0 m * sin(25°)).
Calculating this expression will give you the speed at the bottom of the swing.