What is the minimum speed needed to fire a champagne cork a distance of 11m?
1 answer:
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Answer:
α=9.42 rad/s T= 0.014 rad/s
Explanation:
Low speed= Wi=3 rps = 3*2*π rad/s= 6π rad/sec
Medium speed= Wf= 9 rps= 9*2*π rad/sec= 18 rad/sec
time= t= 4 sec
Angular acceleration=α=
⇒α=(18 π - 6 π)/4
⇒α=9.42 rad/s
b).
Rotational inertia= I = 1.5*10^-3 kg-m^2
Net torque= T = I*α = (1.5*10^-3) * (9.42) rad/s^2
⇒T= 0.014 rad/s^2
5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore
or
Solving for t, we get
We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):