Why Is My Calculated Velocity at the Swing's Apex Incorrect?

[ryho1092]

Homework Statement

The string in the Figure is L = 113.0 cm long and the distance d to the fixed peg P is 91.5 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. How fast will it be going when it reaches the lowest point in its swing?

B-How fast will it be going when it reaches the highest point in its swing?

Homework Equations

Conservation on Energy: TE=PE+KE
KE=1/2 mv^2
PE=mgh or mg(L)---L=length of string

The Attempt at a Solution

I got 4.71 m/s for v for the first part. I got this by using conservation of energy, not moving initally and at the bottom (my reference point) no PE.

I'm stuck on the second part. I found my TE by using the starting point, mgh=PE which is 9.8 m/s^2 * 1.13m * m(mass). This should equal my PEtop + KEtop at the top, so
PEtop = mgh=mg(1.13m-.915m) and KEtop should equal 1/m2v^2. Puting it all together I have the following: TE=PE+KE 9.8m/s^2 * 1.13 m * mass = mass * 9.8 m/s^2 * (1.13-.915)m + 1/2 * mass * v^2. The masses cancle out, and I solve for v= SQRT(2*(9.8*1.13-9.8*1.13-91.5))= 4.23 m/s

But this isn't accepted by capa, what did I do wrong? Thanks in advance

Ryan

 

[SHISHKABOB]

I don't know if you meant to put up a picture, but I'm not really sure what's going on in this problem. And it looks like there's only one part?

 

[ryho1092]

My bad, thanks. I put the second part and attached a pic. Sorry

 

[ryho1092]

I see what I did wrong. I was using the PE at the top not from the pivot point. The EQ for the speed at the top of the pivot is still from TE=PE+KE TE=9.8*1.13*mass PE=9.8*.43*mass the .43 is from 1.14-.915 which gives the radius=.215 r*2=.43 This gives me 6.86 which equals KE=1/2mv^2. Then solve for v and done.

 

[Nickie]

As a scientist, it is important to carefully review your calculations and assumptions to identify any potential errors. In this case, it seems that you may have made a mistake in your calculation for the potential energy at the top of the swing. The correct equation for potential energy at a given height is PE = mgh, where h is the height above the reference point. In this case, the reference point is the bottom of the swing, so the height at the top of the swing would be (L-d), or 113 cm - 91.5 cm = 21.5 cm. This would give a potential energy of PE = mgh = (mass)(9.8 m/s^2)(0.215 m).

Additionally, it seems that you may have made a mistake in your calculation for the kinetic energy at the top of the swing. The correct formula for kinetic energy is KE = 1/2 mv^2, where m is the mass and v is the velocity. In this case, the mass and velocity at the top of the swing would be the same as at the bottom, so KE = 1/2 mv^2 = (mass)(1/2)(v^2).

Combining these corrections, the correct equation would be TE = PE + KE = mgh + 1/2 mv^2. Solving for v, we get v = √(2gh) = √(2(9.8 m/s^2)(0.215 m)) = 1.96 m/s.

It is important to carefully double check your calculations and equations to ensure accuracy. If you are still having trouble, it may be helpful to ask a classmate or instructor for assistance. Keep up the good work!