A uniform horizontal rod of mass 2.4 kg and
length 0.86 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 =
(m*l^2)/12
If a 6.8 N force at an angle of 69� to the hor-
izontal acts on the rod as shown, what is the
magnitude of the resulting angular acceleration about the pivot point? The acceleration
of gravity is 9.8 m/s2 .
Answer in units of rad/s2
I don't understand the force placement.
Use Torque=(momentinertia)angacc
to get the moment of inertia, use the Parallel axis law.
To get torque, force*distance*sinAngle
To find the magnitude of the resulting angular acceleration of the rod, we can use Newton's second law for rotational motion:
τ = Iα
where τ is the torque applied to the rod, I is the moment of inertia of the rod, and α is the angular acceleration.
First, let's calculate the torque applied to the rod. The torque is given by:
τ = r × F
where r is the position vector from the pivot point to the line of action of the force, and F is the force applied.
Since the force is acting at an angle to the horizontal, we need to decompose the force into its horizontal and vertical components. Using trigonometry, we can find the horizontal and vertical components of the force:
F_horizontal = F * cos(θ)
F_vertical = F * sin(θ)
where θ is the angle between the force and the horizontal axis.
Now, we can calculate the torque:
τ = r × F = r * F_vertical
The moment arm, r, is the perpendicular distance from the pivot point to the line of action of the force. In this case, since the force is applied at the center of the rod, the moment arm is equal to the length of the rod divided by 2:
r = l/2
Substituting these values into the torque equation, we have:
τ = (l/2) * F_vertical
Next, let's calculate the moment of inertia of the rod. According to the given formula:
I = (m * l^2) / 12
where m is the mass of the rod, and l is the length of the rod.
Now, to find the angular acceleration, we can rearrange the torque equation:
τ = I * α
α = τ / I
Substituting the values we have calculated, we get:
α = ((l/2) * F_vertical) / ((m * l^2) / 12)
Simplifying the equation, we have:
α = (6 * F_vertical) / (m * l)
Now, let's substitute the given values into the equation:
m = 2.4 kg
l = 0.86 m
F = 6.8 N
θ = 69 degrees
First, we need to convert the angle from degrees to radians:
θ_radians = θ * π / 180
Now, we can calculate the vertical component of the force:
F_vertical = F * sin(θ_radians)
Substituting the values, we get:
F_vertical = 6.8 N * sin(69 * π / 180) ≈ 6.8 N * 0.937
Now, we can substitute the values into the equation for angular acceleration:
α = (6 * F_vertical) / (m * l)
α = (6 * 6.8 N * 0.937) / (2.4 kg * 0.86 m)
Simplifying the equation, we have:
α ≈ 15.874 rad/s^2
Therefore, the magnitude of the resulting angular acceleration about the pivot point is approximately 15.874 rad/s^2.