The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.

Answers

Answer 1

Answer:

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=(0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m)/((\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m) \n\ny=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=(1)/(12)*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\n0.888)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+(1)/(12)*3(1.5)(1.5)^2+\n3(1.5)(0.888-0.75)^2\n\nI_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=(1)/(12)*3(0.8)(0.8)^2+ (1)/(3)* 3(1.5)(1.5)^2+(1)/(2)*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\n(1)/(2)*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\n\nI_o=13.4\ kgm^2$

Answer 2

Answer:

Final answer:

To solve this problem, calculate the mass of each element of the pendulum, use that information to determine the center of mass, and then apply the parallel axis theorem to calculate the two moments of inertia.

Explanation:

To determine the center of mass and the mass moment of inertia of the pendulum, first we calculate the individual masses of the rods: AB and OC, and the plate. Each rod has a mass of 2 kg (given mass per unit length is 3kg/m and length of each rod is 1 m from the first reference paragraph).

The center of mass ӯ can be determined using the formula for center of mass, averaging distances to each mass element weighted by their individual masses. The mass moment of inertia, also known as the angular mass, for rotation about the z axis through G is determined using the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance D away from an axis through the center of mass is the sum of the moment of inertia for rotation about the center of mass and the total mass of the body times D squared.

Finally, the moment of inertia about the z axis passing through the center of rotation O can be calculated again using the parallel axis theorem, with distance d being the distance between points G and O.

Learn more about Mass Moment of Inertia here:

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A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
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Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

$F=(mv^2)/(r)$

$v=\sqrt{(Fr)/(m)}$

$v=\sqrt{(0.025* 0.29)/(0.01)}$

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.Decrease the mass of the car slightly.
Increase the mass of the car slightly.
Decrease the launch speed of the car slightly.
Increase the launch speed of the car slightly.

Answers

Answer:

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

Explanation:

Kinetic energy is

K = ½ m v²

the speed of the expensive we can find it r

v² = v₀² + 2 a x

we can find acceleration with Newton's second law

F = m a

a = F / m

F= cte

substitute in the velocity equation

v² = v₀² + 2 F/m x

let's substitute in the kinetic energize equation

K = ½ m (v₀² + 2 F/m x)

K = ½ m v₀² + f x

we see that the kinetic energy depends on two tomines

in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy

As the launch speed increases the initial energy increases quadratically

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest. The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec. What is the velocity of the first ball after the collision?

Answers

Given :

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest.

The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec.

To Find :

The velocity of the first ball after the collision.

Solution :

We know, by conservation of momentum :

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Putting all given values with directions ( one side +ve and other side -ve ).

$5* 6 + 1* 0 =5 * v_1 + 1* 10\n\n5v_1=10-30\n\nv = -4 \ m/s$

Therefore, the velocity of first ball after the collision is 4 m/s after in opposite direction.

Hence, this is the required solution.

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= (1)/(2)mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m (v^2)/(r)$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2$

1. Hoai Nguyen, a Physics 2A student, drop a soccer ball from the roof of the new science building. The ball strikes the ground in 3.30 s later. You may ignore air resistance, so the ball is in free fall. How tall, in meters, is the building? 2. How fast was the ball moving right before hitting the floor?

3. Thu Tran, another Physics 2A student, grabs the ball and kicks it straight up to Hoai Nguyen, who is still up on the building rooftop. Assuming that the ball is kicked at 0.50 m above the ground and it goes on a vertical path, what is the minimum velocity required for the ball to make it to the building rooftop? Ignore air resistance. (Hint: the ball will pass the rooftop level with a higher speed...)

Answers

Answer:

1. 53.415 m

2. 32.373 m/s

3. 30.82 m/s

Explanation:

Let g = 9.81 m/s2. We can use the following equation of motion to calculate the distance traveled by the ball in 3.3s, and the velocity it achieved

1. $s = gt^2/2 = 9.81*3.3^2/2 = 53.415 m$

2. $v = gt = 9.81*3.3 = 32.373 m/s$

3. If the ball is kicked at 0.5 m above the ground then the net distance between the ball and the roof top is

53.415 - 0.5 = 48.415 m

For the ball to at least make it to the roof top at speed v = 0 m/s. We can use the following equation of motion to calculate the minimum initial speed

$v^2 - v_0^2 = 2g\Delta s$

where v = 0 m/s is the final velocity of the ball when it reaches the rooftop, $v_0$ is the initial velocity, $\Delta s = 48.415$ is the distance traveled, g = -9.81 is the gravitational acceleration with direction opposite with velocity

$0 - v_0^2 = 2*(-9.81)*48.415$

$v_0^2 = 950$

$v_0 = √(950) = 30.82 m/s$

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