Recent content by blue5t1053

Would this be right? m*g*h = \frac{1}{2} * m * v^{2} 9.8 \frac{m}{sec^{2}} * 1 m = \frac{1}{2} * v^{2} \sqrt{\frac{9.8 \frac{m}{sec^{2}} * 1 m}{\frac{1}{2}}} = v v = 4.427 \frac{m}{sec} \ ; \ then *2 m \ for \ top \ of \ rod v = 8.85 \frac{m}{sec}

So if I revisit method one... L_{child} = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3 \frac{kg*m^{2}}{sec} L_{merry-go-round} = (3.91 m) * (371.531 kg) * (0 \frac{m}{sec}) = 0 \frac{kg*m^{2}}{sec} I_{child} = (199 kg) * (3.91 m)^{2} = 3042.33 kg*m^{2} method \ 1 \ \Rightarrow...

I am to assume that the ground is frictionless and that it will rotate from where the rod had made contact with the ground. I can't find any way to calculate it without mass. What principle would I use to find it? I tried using angular velocity, but I couldn't make it work.

I had a lapse of thought there. So I shouldn't include the mass of the merry-go-round?

I thought 'p' was for momentum of the merry-go-round. The only object moving initially is the child. If I apply the child attached to the merry-go-round initially with the starting speed, if I understand this correctly, would be L = (3.91 m) * (199 kg) * (3.09 \frac{m}{sec}) = 2404.3...

L = r *p L = (3.91 m) * (371.53 kg) = 1452 kg*m \ ; \ merry-go-round \ without \ child

Am I to assume that 'm' is total mass of the system (merry-go-round + child) or merry-go-round for interia of the merry-go-round?

Problem: In a plyground there is a small merry-go-round of radius 3.91 m and rotational inertia 5.68e+03 kg m^2. A child of mass 199 kg runs at a speed of 3.09 m/sec tangent to the rim of the merry-go-round when it is at rest and then jumps on. Assume no friction in the bearing of the...

Problem: A rigid assembly is made of a light weight rod 2 m long and a heavy gold ball which is attached at the middle of the rod. The rigid assembly initially rests with one end on the ground in a vertical position. If released what speed does the top of the rod have when it strikes the...

Question: A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1? Hint: let t1=0 sec, and t2=t Equations: \vartheta...

2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec} 3.45575 \frac{rad}{sec} \times .18 m = .622035 \frac{m}{sec} \frac{(.622035 \frac{m}{sec})^{2}}{.18 m} = 2.1496 \frac{m}{sec^{2}} Thank you!

Oh my, I forgot about radius, haha. I was too concerned with the conversion to meters from centimeters!

2 \times \pi \times \frac{.55 rev}{sec} = 3.45575 \frac{rad}{sec} \frac{3.45575 \frac{rad}{sec}}{.36 m} = 9.59931\frac{m}{sec} (9.59931 \frac{m}{sec})^{2} \times .36 m = 33.1728 \frac{m}{sec^{2}} Is this correct?

Problem: What is the linear acceleration of a point on the rim of a 36 cm diameter turntable which is turning at 33 rev/min? My Work: T = \frac{2 \pi r}{v}; v = \frac{2 \pi r}{T} a_{r} = \frac{v^{2}}{r} \frac{33 rev}{min} \times \frac{1 min}{60 sec} = \frac{.55 rev}{sec} \frac{2...

Problem: A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is rotating at 12 rev/s; 69 revolutions later, its angular speed is 33 rev/s. Calculate (a) the angular acceleration, (b) the time required to complete the 69...