Question

PLEASE HELP ME OUT. PLEASE HAVE WITH HANDWRITTEN AND CLEAR HANDWRITTING. PLEASE SHOW ALL YOUR WORK STEP BY STEP SOLUTION. PLEASE. I REALLY NEED CORRECT ANSWER. PLEASE DO ALL PART AS THEY HAVE ASKED TO DO.

Purpose: The purpose of this assignment is for you to demonstrate your physics problem solving skills related to the topics of Rotation of a Rigid Body. Understanding the concepts associated motion in terms of kinematics, dynamics and energy is just the beginning. We can now apply these tools to understanding some specific and practical applications. 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

Answer 1

1) We have assumed that the dinosaur model in the problem is a point object.

The system of rod and the dinosaur model is in equilibrium.

As there is no motion (translational or rotational) of the system of rod and model,

the net external force on the system is zero as well as

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:

--------(i)

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

4)

a) On solving the equation (i)

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

T= (125 * sin 53 * 10) - (4 * 9.81 * ) 10 * 9.81 = 8.18m

Therefore, the distance of the point from the hinge at which the model is hanged is 8.18 m [answer].

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

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

see the diagram:

J30에 니는 mg IN mg

= height traveled by the center of mass of the rod

10 hi = * sin30° L 4 = 2.5m 2

= height traveled by the model

8.18 H2 = r * sin30° = 4.09m 2

Moment of inertia if the system about the hinge is I,

ML2 + m. m.r? 4* 102 -) + (10 * 8.18%) = 802.46kg.m 3 3

Applying conservation of mechanical energy:

loss in potential energy = gain in rotational kinetic energy

1 → Mgh1 + mgh2 1w2 L 21
[where L = angular momentum = $I\omega$ ]

→L= 2(Mgh1 + mgh2)I = V2 * 9.81((4 * 2.5) + (10 * 4.09)) * 802.46

= 895.20 N.m.s

Therefore, the angular momentum after rotation of 30 degree is 895.20 N.m.s [answer]