A bucket crane consists of a uniform boom of mass M = 201 kg and length
L = 59.05 ft that pivots at a point on the bed of a fixed truck. The truck supports an elevated bucket with a worker inside at the other end of the boom, as shown in the figure. The bucket and the worker together can be modeled as a point mass of weight 203 lb located at the end point of the boom.
Suppose that when the boom makes an angle of 69.1 degrees with the horizontal truck bed, the bucket crane suddenly loses power, causing the bucket and boom to rotate freely toward the ground. Find the magnitude of the angular acceleration of the system just after the crane loses power. Take the rotation axis to be at the point where the boom pivots on the truck bed. Use g = 9.81 m/s2 for the acceleration due to gravity. For unit conversions, assume that 1 m = 3.28 ft and 1 lb = 4.45 N. Express your answer to at least two decimal places.
Well, why did the bucket crane lose power? Did someone forget to pay the electricity bill? Or was it just having a moment of existential crisis and decided to take a break? Either way, let's figure out the magnitude of the angular acceleration of this wacky situation.
To solve this problem, we need to apply the principles of torque and rotational motion. The torque acting on an object is equal to the product of the force applied and the distance from the rotation axis.
First, let's convert the given values to SI units because physics loves consistency. The mass of the boom (M) is 201 kg, the length of the boom (L) is 59.05 ft (which is approximately 18 m), and the weight of the bucket and worker (W) is 203 lb (which is approximately 907 N).
When the boom is at an angle of 69.1 degrees with the horizontal, the torque acting on the system is caused by the weight of the bucket and worker. The torque can be calculated using the equation:
Torque = Force × Distance
The force is the weight of the bucket and worker (W), and the distance is the length of the boom (L). So, the torque is:
Torque = W × L
Substituting the given values, we get:
Torque = 907 N × 18 m
To find the angular acceleration (α) of the system, we can use the equation:
Torque = Moment of Inertia × Angular Acceleration
The moment of inertia of a point mass rotating about an axis at a distance (r) is given by:
Moment of Inertia = Mass × Radius^2
In this case, the mass is effectively concentrated at the end point of the boom, so the radius is the length of the boom (L). Thus:
Moment of Inertia = M × L^2
Substituting the given values, we get:
Moment of Inertia = 201 kg × (18 m)^2
Now we can equate torque and moment of inertia:
W × L = M × L^2 × α
Substituting the numerical values, we have:
907 N × 18 m = 201 kg × (18 m)^2 × α
Simplifying this equation will give us the magnitude of the angular acceleration (α). So let's plug it into a calculator or software capable of handling such calculations. Oh, and don't forget to convert the weight given in pounds to newtons before substituting it into the equation.
I hope this answer doesn't make your head spin as much as the crane itself!
To find the magnitude of the angular acceleration of the system just after the crane loses power, we can use the principle of conservation of angular momentum.
The angular momentum of the system before the crane loses power is equal to the angular momentum after the crane loses power.
The angular momentum of the system before the crane loses power is given by:
L_initial = I * ω_initial
where I is the moment of inertia of the system and ω_initial is the initial angular velocity.
The angular momentum of the system after the crane loses power is given by:
L_final = I * ω_final
where ω_final is the final angular velocity.
Since the bucket and the worker inside it can be modeled as a point mass at the end of the boom, the moment of inertia of the system can be calculated as:
I = M * L^2
where M is the mass of the boom and L is the length of the boom.
The weight of the bucket and the worker inside it can be converted to Newtons as follows:
Weight = mass * acceleration_due_to_gravity
Weight = (203 lb) * (4.45 N/lb)
Now, let's calculate the values needed to find the angular acceleration:
1. Convert the length of the boom from feet to meters:
L = 59.05 ft * (1 m / 3.28 ft)
2. Convert the weight of the bucket and worker to Newtons:
Weight = 203 lb * (4.45 N / 1 lb)
Now, we can find the angular velocities before and after the crane loses power:
3. Calculate the initial angular velocity using the angle given:
ω_initial = 69.1 degrees * (π/180 radians/degree)
4. Since the boom and bucket will eventually rotate towards the ground, the final angular velocity can be considered zero:
ω_final = 0 rad/s
Now, let's calculate the angular momentum of the system before and after the crane loses power:
5. Calculate the moment of inertia of the system:
I = M * L^2
6. Calculate the initial angular momentum of the system:
L_initial = I * ω_initial
7. Calculate the final angular momentum of the system:
L_final = I * ω_final
Since angular momentum is conserved, L_initial = L_final:
8. Set L_initial = L_final and solve for the angular acceleration:
I * ω_initial = I * ω_final
9. Substitute the values of I, ω_initial, and ω_final into the equation and solve for the angular acceleration.
Please note that in the provided data, the mass is given in kilograms (kg) and the length is given in feet (ft). To maintain consistent units throughout the calculation, we convert the length from feet to meters. However, for the final answer, we will round it to two decimal places as specified.
To find the magnitude of the angular acceleration of the system just after the crane loses power, we can use the principle of conservation of angular momentum.
The angular momentum of a system is given by the equation:
L = Iω,
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia, I, can be calculated using the formula:
I = (1/3)ML^2,
where M is the mass of the boom and L is the length of the boom.
Given that the mass of the boom, M, is 201 kg and the length of the boom, L, is 59.05 ft, we first need to convert the length from feet to meters:
L = 59.05 ft * (1 m / 3.28 ft) = 17.99 m.
Next, we can calculate the moment of inertia:
I = (1/3) * 201 kg * (17.99 m)^2 = 217808.08 kg·m^2.
Now, we need to find the initial angular velocity, ω_0, just after the crane loses power. To do this, we can use the conservation of angular momentum equation:
L_0 = L_f,
where L_0 is the initial angular momentum and L_f is the final angular momentum.
The initial angular momentum, L_0, can be calculated using:
L_0 = I * ω_0,
where ω_0 is the initial angular velocity.
Given that the weight of the bucket and worker is 203 lb, we first need to convert the weight from lb to Newtons:
Weight = 203 lb * 4.45 N / lb = 903.35 N.
Now, we can calculate the final angular momentum, L_f. At the final angle of 69.1 degrees, the bucket and boom will rotate freely toward the ground. Therefore, the final angular velocity, ω_f, will be zero.
L_f = (203 lb * 4.45 N / lb) * (17.99 m),
L_f = 162431.85 N·m·s.
Setting L_0 equal to L_f, we have:
I * ω_0 = L_f,
(217808.08 kg·m^2) * ω_0 = 162431.85 N·m·s.
Solving for ω_0, we find:
ω_0 = 162431.85 N·m·s / (217808.08 kg·m^2) = 0.7468 rad/s.
Finally, we can find the angular acceleration, α, using the equation:
α = (ω_f - ω_0) / Δt,
where ω_f is the final angular velocity (zero in this case) and Δt is the time taken for the system to come to a stop.
Since the system comes to a stop instantly after the crane loses power, Δt can be considered to be zero. Therefore, the angular acceleration, α, is also zero.
Hence, the magnitude of the angular acceleration of the system just after the crane loses power is zero.
moment of inertia of boom about pivot = (1/3) m L^2
= (1/3)(201)(59.05^2)
moment of inertia of man/bucket about pivot = (203 lb *4.45 N/lb)(59.05^2/3.28^2)
add them for total I
Torque of arm about pivot = (201 kg *9.81) (59.05/3.28) (cos 69.1) /2
Torque of man/bucket about pivot = (203*4.45)(59.05/3.28) (cos 69.1)
add them for total torque
a = torque / I