Calculating Speed of Meterstick Swinging Vertical - Physics

[lina29]

A stick with a mass of 0.152 kg and a length of 1.00 m is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released.
C-As it swings through the vertical, calculate the linear speed of the end of the stick opposite the axis.
D-Find the ratio of the speed of a particle that has fallen a distance of 1.00 m, starting from rest, to the speed from part (C).
For part C I got 5.42 m/s I just cannot figure out part d. Any help would be greatly appreciated!

 

[berkeman]

A stick with a mass of 0.152 kg and a length of 1.00 m is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released.
C-As it swings through the vertical, calculate the linear speed of the end of the stick opposite the axis.
D-Find the ratio of the speed of a particle that has fallen a distance of 1.00 m, starting from rest, to the speed from part (C).
For part C I got 5.42 m/s I just cannot figure out part d. Any help would be greatly appreciated!

Welcome to the PF.

Please show us your work for how you got part (C).

 

[lina29]

v=rw= (1 m)(5.42 rad/s)=5.42 m/s

 

[lina29]

Never mind I didn't read the problem correct. I didn't realize it said ratio :)
Thanks anyway!

 

[rwisz]

For part D, we can use the conservation of energy principle. At the initial position, the stick has potential energy due to its height above the ground. As it swings down, this potential energy is converted into kinetic energy. Therefore, we can equate the initial potential energy to the final kinetic energy:

mgh = 1/2 mv^2

Where m is the mass of the stick, g is the acceleration due to gravity, h is the height from which the particle falls, and v is the final velocity.

Solving for v, we get:

v = √(2gh)

Substituting the values, we get:

v = √(2*9.8*1.00) = 4.43 m/s

Therefore, the ratio of the speed of the particle to the speed of the end of the stick opposite the axis is:

4.43/5.42 = 0.82

This means that the particle has a speed that is 82% of the speed of the end of the stick.