A uniform horizontal rod of mass 2.2 kg and length 0.63 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I =
ml2/12
If an 8.5N force at an angle of 47◦ to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular accel- eration about the pivot point? The accelera- tion of gravity is 9.8 m/s2 . Answer in units of rad/s2
Angular acceleration = (Torque)/I
They have given you the formula for I:
M L^2/12
I would need to see the figure to tell you what the torque about the spin axis is.
To find the magnitude of the resulting angular acceleration, we can use the torque equation:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
First, let's determine the torque caused by the 8.5N force. The torque is given by the cross product of the force and the lever arm.
τ = r * F * sin(θ)
where r is the lever arm (distance from the pivot point to the line of action of the force), F is the magnitude of the force, and θ is the angle between the force and the lever arm.
In this case:
r = 0.63 m (since the force is applied at the end of the rod)
F = 8.5 N
θ = 47°
We need to convert θ to radians:
θ (in radians) = (47° * π) / 180°
Now we can calculate the torque:
τ = (0.63 m) * (8.5 N) * sin((47° * π) / 180°)
Next, we can calculate the moment of inertia of the rod about the pivot point:
I = (m * l^2) / 12
where m is the mass of the rod and l is the length of the rod.
In this case:
m = 2.2 kg
l = 0.63 m
I = (2.2 kg * (0.63 m)^2) / 12
Now we can substitute these values into the torque equation to find the angular acceleration:
τ = I * α
(0.63 m) * (8.5 N) * sin((47° * π) / 180°) = ((2.2 kg * (0.63 m)^2) / 12) * α
Solving for α:
α = [(0.63 m) * (8.5 N) * sin((47° * π) / 180°)] / ((2.2 kg * (0.63 m)^2) / 12)
Now we can calculate the value of α.
To determine the magnitude of the resulting angular acceleration, we can use Newton's second law for rotational motion, which states that the net torque on an object is equal to the moment of inertia times the angular acceleration.
The net torque acting on the rod is caused by two forces: the gravitational force and the applied force. The gravitational force acting on the rod can be calculated using the formula F = mg, where m is the mass of the rod and g is the acceleration due to gravity.
F_gravity = mg
= (2.2 kg)(9.8 m/s^2)
= 21.56 N
The torque due to the gravitational force is given by τ_gravity = F_gravity * d, where d is the distance from the pivot point to the center of mass. Since the rod is uniform, the center of mass is located at the midpoint of the rod, which is equal to half of its length.
d = l/2
= 0.63 m / 2
= 0.315 m
τ_gravity = F_gravity * d
= (21.56 N)(0.315 m)
= 6.79 Nm
Now, let's calculate the torque due to the applied force. The torque can be found using the formula τ = F * r * sin(θ), where F is the magnitude of the force, r is the distance from the pivot point to the point at which the force is being applied (perpendicular to the line of action of the force), and θ is the angle between the force and a line drawn from the pivot point to the point of application.
r = l
= 0.63 m
τ_applied = F_applied * r * sin(θ)
= (8.5 N)(0.63 m) * sin(47°)
= 5.34 Nm
The net torque is the sum of the torques due to the gravitational force and the applied force.
τ_net = τ_gravity + τ_applied
= 6.79 Nm + 5.34 Nm
= 12.13 Nm
Now, we can find the angular acceleration using the formula τ = I * α, where I is the moment of inertia and α is the angular acceleration.
Let's substitute the given moment of inertia, I = ml^2/12.
τ_net = I * α
12.13 Nm = (2.2 kg)(0.63 m)^2 / 12 * α
Now, solve for α.
α = 12.13 Nm / [(2.2 kg)(0.63 m)^2 / 12]
≈ 79 rad/s^2
Therefore, the magnitude of the resulting angular acceleration about the pivot point is approximately 79 rad/s^2.