Find Speed of Cylinder Rolling Down Loop | m=0.710 kg, h=333 cm, R=49.95 cm

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The discussion centers on calculating the speed of a uniform solid cylinder rolling down a ramp and around a loop. The initial approach used conservation of energy, equating potential energy at the top of the ramp to kinetic energy at the top of the loop. The user initially calculated the speed as 6.60 m/s, but recognized that the height used should be the difference between the ramp's height and the loop's height. After clarifying this point, the user found the correct approach to solve the problem. The thread emphasizes the importance of accurately determining the height in energy conservation calculations.
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A uniform solid cylinder (m=0.710 kg, of small radius) is at the top of a ramp, with height height of 333 cm and which has a loop of radius, R = 49.95 cm at the bottom. Which has friction. The cylinder starts from rest and rolls down the ramp without sliding and goes around the loop. Find the speed of the cylinder at the top of the loop.

I used conservation of energy
mgh= (1/2)mv^2 + (1/2)I \omega ^2
mgh = (1/2)mv^2 + (1/2)(1/2)(MR^2)(v/r)^2
mgh= (1/2)mv^2 +(1/4)mv^2
Solving for v gave me \sqrt (4/3)gh
So plugging in 9.8 for g and 3.33 for h gave me a speed of 6.60 m/s which wasn't right.
Can someone help me?
 
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h should be the difference in height between the top of the ramp and the top of the loop.
 
I got it. Thanks
 
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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