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4 years ago

What is the moment of inertia i of this assembly about the axis through which it is pivoted?

Physics

1 answer:

vivado [14]4 years ago

7 0

Answer:

I = $\frac{1}{12}m_t(2x)^2+m_1x^2+m_2x^2$

Explanation:

The moment of inertia for the beam is:

I = $\frac{1}{12}m_tL^2$

Where $m_t$ is the mass of the beam and L is the lengh of the beam

note:

L = 2x

And for particles I is equal to:

I = MR^2

where M is the mass of the particle and R is the distance between the pivot and the particle.

Finally, the moment of inertia for this assembly is the sum of the moment of inertia of the particles and the beam. So:

I = $\frac{1}{12}m_t(2x)^2+m_1x^2+m_2x^2$

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An auto race takes place on a circular track. A car completes one lap in a time of 25.0 s, with an average tangential speed of 5

ludmilkaskok [199]

Answer:

(a) Angular speed will be 0.2512 rad/sec

(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = $2\pi$ radian

(a) So angular speed $=\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec$

(B) Tangential velocity is given as v = 53.6 m/sec

We know that tangential velocity is given by

$v=\omega r$

So radius $r=\frac{v}{\omega }=\frac{53.6}{0.2512}=213.375m$

4 0

3 years ago

Suppose the experiment was conducted in the same manner, but the axle was now off center of the solid disk. Would you expect the

san4es73 [151]

Answer:

Explanation:

If Ig be moment of inertia about an axis through centre of mass and I be moment of inertia through any other axis parallel to earlier axis , then according to theory of parallel axis ,

I = Ig + Md²

where M is mass of the body and d is distance between two parallel axis.

So I is greater than Ig.

4 0

3 years ago

How does the electric field intensity vary with the increase of distance of the point from the centre of a charged conducting sp

Nataly_w [17]

Answer:

(i) Electric field outside the shell:

For point r>R; draw a spherical gaussian surface of radius r.

Using gauss law, ∮E.ds=q0qend

Since E is perpendicular to gaussian surface, angle betwee E is 0.

Also E being constant, can be taken out of integral.

So, E(4πr2)=q0q

So, E=4πε01r2q

4 0

3 years ago

Arrange the objects in order from least to greatest potential energy. Assume that gravity is constant. (Use PE = m × g × h.)

yawa3891 [41]

Answer:The objects in order from least to greatest potential energy:

3<1<4<2

Explanation:

1. Potential energy of a 15-kilogram stone found at a height of 3 meters:

$mgh=15 kg\times 9.8 m/s^2\times 3 m=441 J$

2. Potential energy 10 kilograms of water stored at a height of 9 meters:

$mgh=10 kg\times 9.8 m/s^2\times 9 m=882 J$

3. Potential energy 1-kilogram ball located 20 meters in the air:

$mgh=1 kg\times 9.8 m/s^2\times 20 m=196 J$

4. Potential energy box of books weighing 25 kilograms placed on a shelf 2 meters high:

$mgh=25 kg\times 9.8 m/s^2\times 2 m=490 J$

The objects in order from least to greatest potential energy:

3<1<4<2

8 0

3 years ago

Read 2 more answers

In a carnival game, the player throws a ball at a haystack. For a typical throw, the ball leaves the hay with a speed exactly on

8_murik_8 [283]

Answer:

Ve(m) = sqrt (19.2/m)

Ve(0.35) = 7.407 m/s

Explanation:

Given:

- The ball has a mass = m

- The entry speed of the ball is Vi = Ve

- The final speed of the ball Vf = 0.5*Ve

- The constant frictional force on ball due to hay is F = 6 N

- The thickness of hay-stack is s = 1.2 m

- Assume the throw is in horizontal direction and neglect gravity forces

Find:

Derive an expression for the typical entry speed as a function of the inertia of the ball

What is the typical entry speed if the ball has an inertia of a 0.35 kg?

Solution:

- To determine the entry speed as a function of inertia we will use third equation of motion as follows:

Vf^2 = Vi^2 + 2*a*s

Where, a is acceleration of the ball through hay stack. We will use Newton's Law of motion to determine this:

F_net = m*a

The only force acting on the ball in its journey through hay-stack is the frictional force F:

- F = m*a

a = -F/m

- Input all the quantities in the third equation of motion:

(0.5Ve)^2 = Ve^2 - 2*F*s / m

0.75Ve^2 = 2*F*s / m

Ve = sqrt (8*F*s/3*m)

Plug in values:

Ve(m) = sqrt (8*6*1.2/3*m)

Ve(m) = sqrt (19.2/m)

- The entry speed for the inertia of the ball m = 0.35 kg is:

Ve(0.35) = sqrt(19.2/0.35)

Ve(0.35) = 7.407 m/s

8 0

4 years ago