please show all work 3. A rod with a mass of 10kg and a total length of 2 meters that can spin about an axis through itscenter ( I = ML2 ) receives a constant force of 5 N on one of its ends. As the rod spins, the12force will maintain the same strength and orientation i.e., angled at 60 degrees with respect to therod. The rod is initially at rest.Total LengthAxis60 degF,a. How many radians does the rod sweep through after a period of 10 seconds?b. What is the speed of a point on the end of the rod at that same moment?

Answered: b. What is the speed of a point on the…

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please show all work

3. A rod with a mass of 10kg and a total length of 2 meters that can spin about an axis through its
center ( I = ML2 ) receives a constant force of 5 N on one of its ends. As the rod spins, the
12
force will maintain the same strength and orientation i.e., angled at 60 degrees with respect to the
rod. The rod is initially at rest.
Total Length
Axis
60 deg
F,
a. How many radians does the rod sweep through after a period of 10 seconds?
b. What is the speed of a point on the end of the rod at that same moment?

Expert Solution

Step 1

Moment of Inertia

The moment of inertia of a rigid body about an axis is the sum of the product of the masses of the particles constituting the body and the square of the distance between the particles and the axis of rotation. Let the mass of $n$ particles be $m_{1}, m_{2}, . . ., m_{n}$ and the distance between the particles and the axis of rotation is $r_{1}, r_{2}, . . ., r_{n}$ , then the moment of inertia is given by

$I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + . . . + m_{n} r_{n}^{2} = \sum_{i = 1}^{n} m_{i} r_{i}^{2}$

The torque acting on a body is equal to the cross product of the force acting on the body and the shortest perpendicular distance between the force and the axis of rotation.

$\vec{τ} = \vec{F} \times \vec{d} = F d \sin θ \hat{n}$

In rigid body dynamics, the total external torque acting on the system is equal to the product of the moment of inertia and the angular acceleration.

$τ = I α$

The equations of motion followed in rotational dynamics are similar to those of rectilinear motion

$ω_{f}^{2} = ω_{i}^{2} + α t θ = ω_{i} t + \frac{1}{2} α t^{2} ω_{f}^{2} = ω_{i}^{2} - 2 α θ$

The relation between the angular velocity $ω$ and linear velocity $v$ is

$v = ω R$

$R$ is the distance between the point of interest and the center of rotation.

Step 2

In the given question force always makes an angle of $60 °$ with the rod. Therefore if the rod is rotating along an axis perpendicular to the rod and passing through its center then $d = \frac{L}{2} = \frac{2}{2} m = 1 m$

Therefore total external torque acting on the rod

$τ = F d \sin 60 ° = 5 \times 1 \times \sin 60 ° = 4.33 N m$

The Moment of inertia of the rod about the axis passing through the center

$I = \frac{M L^{2}}{12} = \frac{10 \times 2^{2}}{12} k g m^{2} = 3.33 k g m^{2}$

Therefore we have

$τ = I α \Rightarrow 4.33 = 3.33 \times α \Rightarrow α = \frac{4.33}{3.33} = 1.299 r a d / s^{2}$

a. The rod was initially at rest. Therefore initial angular velocity is zero. Using equation motion

$θ = ω_{i} t + \frac{1}{2} α t^{2}$

We can find the angular displacement after $t = 10 s$

$θ = \frac{1}{2} \times 1.299 \times 10^{2} = 64.95 r a d$

b. We can calculate the angular velocity after $t = 10 s$ using the equation

$ω_{f} = ω_{i} + α t$

This gives

$ω_{f} = 1.299 \times 10 r a d / s = 12.99 r a d / s$

Therefore the linear velocity at that moment

$\begin{array}{rcl} v & = & ω_{f} R = ω_{f} \times \frac{L}{2} \\ = & 12.99 \times \frac{2}{2} m / s = 12.99 m / s \end{array}$

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