A uniformrodofmass2kgandlength1mis supportedby a frictionlesspivot 25 cm down fromthetop,asshown.A smallballof mass 0.5 kg strikesthe stationaryrod horizontallyat 2 m/sandsticks. Astherodswings,whatisthe maximumanglee in degreesthatit makeswith the vertical?
To find the maximum angle that the rod makes with the vertical, we can apply the principles of conservation of momentum and conservation of angular momentum.
First, let's consider the initial momentum of the system. The small balloon, with a mass of 0.5 kg, strikes the stationary rod horizontally at 2 m/s. Since the rod is at rest initially, its momentum is zero. Therefore, the total initial momentum of the system is equal to the momentum of the balloon, which is given by:
Initial momentum = mass of the balloon x velocity of the balloon
= 0.5 kg x 2 m/s
= 1 kg m/s
Now, let's consider the final state of the system when the rod swings. At the maximum angle, the rod will momentarily come to rest before reversing its direction. At this point, the angular momentum of the system is conserved.
The angular momentum of the system is given by:
Angular momentum = moment of inertia x angular velocity
Since the rod is a uniform rod rotating about a pivot, its moment of inertia can be calculated using the following formula:
Moment of inertia = (1/3) x mass x length^2
Plugging in the values:
Moment of inertia = (1/3) x 2 kg x (1 m)^2
= 2/3 kg m^2
Now, let's assume the final angular velocity of the rod is ω, and the distance of the center of mass from the pivot is d = 25 cm = 0.25 m.
Then, the moment of inertia of the rod about the pivot is given by:
I = m x d^2
Plugging in the values:
I = 2 kg x (0.25 m)^2
= 0.125 kg m^2
Since angular momentum is conserved, we have:
Initial angular momentum = Final angular momentum
m x v x d = I x ω
Plugging in the values:
1 kg m/s x 0.25 m = 0.125 kg m^2 x ω
Simplifying the equation:
0.25 kg m^2/s = 0.125 kg m^2 x ω
ω = (0.25 kg m^2/s) / (0.125 kg m^2)
ω = 2 rad/s
Now, we can find the maximum angle the rod makes with the vertical using the relationship between linear and angular velocity:
v = ω x r
where v is the linear velocity of the center of mass of the rod and r is the distance of the center of mass from the pivot.
Plugging in the values:
2 m/s = 2 rad/s x 0.25 m
2 m/s = 0.5 m/s
Finally, we can find the maximum angle using the formula:
θ = arcsin(v/r)
Plugging in the values:
θ = arcsin(0.5 m/s / 0.25 m)
θ = arcsin(2)
Calculating the maximum angle using a calculator or trigonometric table, we find:
θ ≈ 63.4 degrees
Therefore, the maximum angle the rod makes with the vertical is approximately 63.4 degrees.