Question

Task, to be written out by hand: 1) Read the following problem and identify the relevant physical principles and summarize them in words. 2) Draw a diagram for the object. Include all the forces, the distance from pivot point they are acting and the angle that each is being applied. 3) Write out the rotational equivalent of Newton's Second Law as it pertains to this problem. 4) Answer the questions listed below. Please do all the necessary algebra before plugging in the numerical values. Provide a final answer with the correct number of significant digits, the appropriate units and an explanation. Grading Rubric: This assignment will be graded out of a total of 20 points. Each task is worth 5 points. Problem: A 10 kg dinosaur sign is to be hung from a long thin rod. The rod is attached to the wall by a hinge at point P and has a cable at the far end attached to the wall at a height h above P. The rod is 10.0 m long and held horizontal by the cable that makes an angle with the rod of 53 degrees. It also has a mass of 4.0 kg. a) How far along the rod should the sign be fastened such that the tension in the cable is to be 125 N (relative to the wall)? b) If the cable was accidentally cut what would the angular momentum of the system be when the rod was 30 degrees below the horizontal before smashing into Caroline's Dino-Store window (model the sign as a particle)? Caroline's Dino-Store w

I need help with 1,2,3,4a and 4b

Answer 1

1) The system of rod and the model is in equilibrium.

The net external force on the system is zero

The net external torque on the system is zero.

When the cable is cut, the only force that performs work is gravity, therefore, we can apply the conservation of mechanical energy.

2) See the diagram:

H-23-e, Tisel 42 Tune ५ M W - L

M = mass of the rod

m = mass of the model

x = distance of the force due to gravity on the model.

= horizontal component of the hinge force

H.,

= vertical component of the hinge force

H y

3) Applying the rotational equivalent of Newton's 2nd Law.

Net torque = moment of inertia about the axis of rotation * angular acceleration

The system is in rotational equilibrium, so the angular acceleration is zero.

As the net force is zero, the net torque about any point in the plane of the diagram is same.

Therefore, balancing the torque of the forces about the hinge:

$(T\sin\theta*L) - (Mg*\frac{L}{2})-(mg*x)=0$--------(i)

4)

a) On solving the equation (i)

T = (T sin 0* L) - (Mg* mg

$\Rightarrow x=\frac{(125*\sin53^{\circ}*10) - (4*9.81*\frac{10}{2})}{10*9.81}=8.18m$ [answer]

b) After the cable cut, the rod makes an angle 30 degree below horizontal.

Let the angular speed at this position be $\omega$ .

see the diagram:

J30에 니는 mg IN mg

$h_{1}=\frac{L}{2}*sin30^{\circ}=\frac{L}{4}=\frac{10}{4}=2.5 m$

$h_{2}=x*sin30^{\circ}=\frac{x}{2}=\frac{8.18}{2}=4.09 m$

Moment of inertia if the system about the hinge = $I = \frac{ML^{2}}{3}+mx^{2} = (\frac{4*10^{2}}{3})+(10*8.18^{2})=802.46kg.m^{2}$

Applying conservation of mechanical energy:

loss in potential energy = gain in rotational kinetic energy

$\Rightarrow Mgh_{1}+mgh_{2} = \frac{1}{2}I\omega^{2} = \frac{L^{2}}{2I}$ [where L = angular momentum = $I\omega$ ]

$\Rightarrow L = \sqrt{2(Mgh_{1}+mgh_{2})I}=\sqrt{2*9.81((4*2.5)+(10*4.09))*802.46}$

= 895.20 N.m.s [answer]