What is the angular displacement of a rolling ball falling from a height?

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To calculate the angular displacement of a rolling ball falling from a height, first determine the time it takes to fall 2.30 m, which is approximately 0.68 seconds. The ball maintains a constant linear speed of 3.00 m/s horizontally, so its angular velocity can be calculated using the formula ω = v/r, resulting in an angular velocity of about 11.54 rad/s. During the time in the air, the ball rolls without slipping, leading to an angular displacement of approximately 7.85 radians. It's important to note that while the vertical speed changes during the fall, the horizontal speed and angular velocity remain constant. The calculations demonstrate the relationship between linear and angular motion for a rolling ball.
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A ball of radius 0.260 m rolls along a horizontal table top with a constant linear speed of 3.00 m/s. The ball rolls off the edge and falls a vertical distance of 2.30 m before hitting the floor. What is the angular displacement of the ball while it is in the air? (assume that there is no slipping of the surfaces in contact during the rolling motion.)
 
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What's the angular velocity of the ball? (You can calculate this, because you know its linear speed and radius).

What is the amount of time the ball is in the air?

Show us what you have done.
 
Also note that only the vertical speed changes while the ball is falling, not the horizontal speed or angular velocity.
 
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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