Conservation of Mechanical Energy for a Rolling Sphere

AI Thread Summary
The discussion focuses on a physics problem involving a solid sphere rolling without slipping on an incline. The total energy of the sphere is a combination of its translational and rotational kinetic energy, which can be calculated using the sphere's mass and speed. The relationship between translational and rotational velocity is essential, as it is determined by the rolling condition. The conservation of mechanical energy principle indicates that the initial energy will equal the potential energy at the maximum height on the incline. Understanding the rotational inertia of a solid sphere is crucial for solving the problem accurately.
adstroud
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I have one last question due on my physics homework that is due in a few and no one seems to understand how to do it. Please help :)


A solid sphere of mass 0.595 kg rolls without slipping along a horizontal surface with a translational speed of 5.16 m/s. It comes to an incline that makes an angle of 36 with the horizontal surface. Neglecting energy losses due to friction,

(a) what is the total energy of the rolling sphere?
Im pretty sure that this is a total of the trans. velocity and rotational velocity but I don't know how to get the rotational velocity from the information given.

(b) to what vertical height above the horizontal surface does the sphere rise on the incline?
 
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adstroud said:
(a) what is the total energy of the rolling sphere?
Im pretty sure that this is a total of the trans. velocity and rotational velocity but I don't know how to get the rotational velocity from the information given.
The key is that it rolls without slipping. That should tell you the relationship between translational and rotational velocity.

What's the rotational inertia of a solid sphere?
 
this should help:

K.E= 1/2 mv^2 ( 1+ k^2/r^2)
 
so what is the vertical height it goes on the incline
 
adstroud said:
so what is the vertical height it goes on the incline
Hint: Mechanical energy is conserved.
 
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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