How Does Pendulum Speed Change with Angle and Initial Velocity?

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The discussion revolves around calculating the speed of a pendulum bob at its lowest position and determining the minimum initial velocity required for the pendulum to swing to a horizontal position. The calculated speed at the lowest point with an initial speed of 8 m/s is 8.35 m/s. For the pendulum to reach the horizontal position, the minimum initial speed required is 4.33 m/s, as the final kinetic energy will be zero at that point. Participants clarify that the value of 8.35 m/s is not applicable for the second part of the question. The conversation concludes with a user expressing confidence in solving the problem after a brief pause.
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[SOLVED] Pendulum Momentum Question

Homework Statement



Figure 8-34 shows a pendulum of length L = 1.25 m. Its bob has speed v0 when the cord makes an angle theta0 = 40 degrees with the vertical.

What is the speed of the bob when it is in its lowest position if v0 = 8 m/s?

The speed is 8.35 m/s.

What is the least value that v0 can have if the pendulum is to swing down and then up to a horizontal position?

Homework Equations



W + KE1 + PE1 = KE2 + PE2

The Attempt at a Solution



W + KE1 + PE1 = KE2 + PE2:

0 + (1/2)m(8.57 m/s)^2 + 0 = (1/2)m v^2 + m(9.8 m/s^2)1.25m

v= 6 or 7 something. The answer in the book is 4.33 m/s
 
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Shackleford said:

Homework Statement



Figure 8-34 shows a pendulum of length L = 1.25 m. Its bob has speed v0 when the cord makes an angle theta0 = 40 degrees with the vertical.

What is the speed of the bob when it is in its lowest position if v0 = 8 m/s?

The speed is 8.35 m/s.

What is the least value that v0 can have if the pendulum is to swing down and then up to a horizontal position?

Homework Equations



W + KE1 + PE1 = KE2 + PE2

The Attempt at a Solution



W + KE1 + PE1 = KE2 + PE2:

0 + (1/2)m(8.57 m/s)^2 + 0 = (1/2)m v^2 + m(9.8 m/s^2)1.25m

v= 6 or 7 something. The answer in the book is 4.33 m/s

Use 8.35 m/s.
 
physixguru said:
Use 8.35 m/s.

Sorry. I forgot to put the 8.35 in there. But, it still doesn't give me the correct answer. I get 6.74 m/s.
 
The final kinetic energy will be zero when the pendulum reaches the horizontal when considering the minimum initial velocity.

The 8.35m/s does not apply here, since that value is based on the first part of the question and is irrelevant to this part. You are trying to find the velocity of the pendulum needed at 40 degrees to the vertical that will just get you to the horizontal position on the other side.
 
hage567 said:
The final kinetic energy will be zero when the pendulum reaches the horizontal when considering the minimum initial velocity.

The 8.35m/s does not apply here, since that value is based on the first part of the question and is irrelevant to this part. You are trying to find the velocity of the pendulum needed at 40 degrees to the vertical that will just get you to the horizontal position on the other side.

Oh, hello! Man. I know I can work it now. I'll do it after I eat. lol.
 
Got it. Thanks. Sometimes, I just read over the problem too quickly and assume I read it correctly.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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