Find Max Speed for Smaller Mass in Trebuchet w/ Stiff Rod & 2 Masses

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In analyzing the trebuchet system with a stiff rod and two masses, the initial potential energy of the masses is set to zero. The larger mass's potential energy loss converts into kinetic energy for both masses, as it does not stop at the bottom but oscillates. The equations used to estimate the velocities must account for this energy transfer, emphasizing the relationship between the tangential speeds of the masses. The assumption that the larger mass stops moving at the bottom is incorrect and leads to discrepancies in calculated speeds. A proper energy analysis reveals the correct maximum speed for the smaller mass.
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Imagine a trebuchet with a stiff rod of 3m and neligible mass.Two masses 60kg and 0.12kg are at its end with the bigger mass 0.14m away from the pivot point. Find the maximum speed the smaller mass attain.

Actually i posted this before but it seems i can't get the answer provided in the book. So what i did was to assume the initial position of the masses to be at 0 potential so initially they have no energy. Then i consider the case when the rod is vertical which gives me the equation - MgL+ \frac{1}{2}Mv12 - mg(0.14) = 0. In this case i assume the bigger mass to stop moving when it reaches the bottom. Am i allowed to make this assumption? If not how do i calculate for the velocity of the bigger mass?

Since the smaller mass is significantly smaller than the bigger mass, i use mg(0.14) - \frac{1}{2}mv22=0 to estimate the velocity of the bigger mass and find the angular velocity from there. I use v=r\omega to calculate the velocity of the smaller mass however the answer i got was way too smaller compared to the answer why is this so?
 
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semc said:
Imagine a trebuchet with a stiff rod of 3m and neligible mass.Two masses 60kg and 0.12kg are at its end with the bigger mass 0.14m away from the pivot point. Find the maximum speed the smaller mass attain.

Actually i posted this before but it seems i can't get the answer provided in the book. So what i did was to assume the initial position of the masses to be at 0 potential so initially they have no energy. Then i consider the case when the rod is vertical which gives me the equation - MgL+ \frac{1}{2}Mv12 - mg(0.14) = 0. In this case i assume the bigger mass to stop moving when it reaches the bottom. Am i allowed to make this assumption? If not how do i calculate for the velocity of the bigger mass?

Since the smaller mass is significantly smaller than the bigger mass, i use mg(0.14) - \frac{1}{2}mv22=0 to estimate the velocity of the bigger mass and find the angular velocity from there. I use v=r\omega to calculate the velocity of the smaller mass however the answer i got was way too smaller compared to the answer why is this so?
You are not taking the correct approach. You have to analyse the problem from the point of view of energy.

The energy of the falling large mass does three things. The loss of potential energy of the large mass gives the large mass kinetic energy; it gives the smaller mass kinetic energy; and it does something else that uses energy. (hint: What happens to the potential energy of the small mass as the large mass falls?). The large mass does not stop when it reaches the bottom. It oscillates until it gradually stops.

When you determine how much energy is kinetic energy, you can determine the speeds. What is the relationship between the tangential speeds of the masses?

AM
 
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