Does a Vertical Force Affect Horizontal Speed on a Frictionless Surface?

  • Thread starter Thread starter gunblaze
  • Start date Start date
AI Thread Summary
A vertical force applied to a ball rolling on a frictionless surface does not affect its horizontal speed. The ball maintains its horizontal velocity regardless of vertical forces, as there is no friction to influence its motion. If the initial force was not directed through the center of mass, the ball may still roll, but this does not alter its horizontal speed. The discussion emphasizes that the conditions of a frictionless environment are crucial to understanding the motion. Overall, vertical forces do not impact horizontal velocity on a frictionless surface.
gunblaze
Messages
187
Reaction score
0
Lets say i have a ball rolling horizontally at constant speed on a frictionless table in vacumm. When i exert a constant vertical force on the ball. Will the horizontal speed change due to the resultant force in the vertical direction? thx
 
Physics news on Phys.org
No. As long as its a frictionless surface, a vertical force won't result in any horizontal velocity change.
 
If its frictionless the ball won't be rolling.
 
whozum said:
If its frictionless the ball won't be rolling.
If the force that started the movement was not applied in a direction passing through the center of mass, the ball will roll even in a frictionless surface.
 
SGT said:
If the force that started the movement was not applied in a direction passing through the center of mass, the ball will roll even in a frictionless surface.

Only if the angular velocity was the exact amount such that for every 2pi radians swept the ball traveled 2\piR units of length.. which is way more assuming than necessary for the problem.
 
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}...
Back
Top