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lapo4ka [179]
4 years ago
11

Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. otherwise, we keep

the experiment the same. compared to the pucks in the video, this pair of pucks will rotate
Physics
2 answers:
Mrac [35]4 years ago
6 0

The pucks will rotate with the same angular rate even after changing the masses of the puck.

Further Explanation:

The two hover pucks collide with each other and stick to one another after the collision. Use the concept of momentum conservation in order to obtain the final rate of rotation of the pucks.

The conservation of momentum means that the momentum of the two bodies remains equal to the momentum of the combined two bodies after they collide if there is no other means of energy loss.

Concept:

Let the two hover pucks have masses {m_1} and {m_2} which are being rotated at the rate of {v_1} and {v_2} respectively.

We can write the expression for the momentum conservation of the pucks as.

\fbox{\begin\\{m_1}{v_1}+{m_2}{v_2}=\left({{m_1}+{m_2}}\right)v\end{minispace}}                        …… (1)

Here, {m_1} is the mass of first puck, {m_2} is the mass of the second puck, {v_1} is the linear rate of the first puck, {v_2} is the linear rate of the second puck and v is the final rate of the two hover pucks.

The rate of the rotation of the pucks after collision will be,

v = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}

Now, when the masses of the two pucks is doubled and the other conditions are kept same.

Substitute 2{m_1} for {m_1} and 2{m_2} for {m_2} in above expression.

\begin{aligned}v'&=\frac{{2{m_1}{v_1}+2{m_2}{v_2}}}{{2{m_1}+2{m_2}}}\\&=\frac{{{m_1}{v_1}+{m_2}{v_2}}}{{{m_1}+{m_2}}}\\&=v\\\end{gathered}

Thus, the above expression shows that the momentum of the hover pucks remains conserved and the hover pucks rotate with the same angular rate.

Learn More:

1.  Collision of a car with the wall brainly.com/question/9484203

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Net force on a body brainly.com/question/4033012

Answer Details:

Grade: High School

Subject: Physics

Chapter: Conservation of Momentum.

Keywords:

Hover, pucks, rotate, momentum, conservation, angular rate, rate of rotation, collide, masses, collision, m1v1, M1V1, (m1+m2)v, m2v2, first puck, second puck, same rate.

kvv77 [185]4 years ago
5 0

This pair of pucks will rotate at the same rate.

Further Explanation:

Hover Puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

Employments of hover puck:  

The Hover Puck coasts effectively on a self-produced pad of air until followed up on by an outside power.  

• Use one Hover Puck to show themes from Newton's First Law to impacts to reflection.  

• Use at least two pucks to examine the protection laws in two measurements.  

• Set up a bowling alley utilizing plastic beverage bottles.  

Innovation of hover puck:  

The First Rubber Hockey Pucks Were Made From Sliced-Up Lacrosse Balls. At the point when the game moved inside, entire balls were initially utilized, yet arena proprietors before long thought that it was desirable over cut them into thirds and keep the center segment. This fundamental plan was the standard by 1885.  

hockey puck was developed:  

The hockey puck appeared in 1875. It's hazy who really imagined it. Specialists accept the main hockey puck was likely only an elastic ball cut down the middle. This gave players an article with a level side that would slide over the ice.

Subject: physics

Level: High School

Keywords: Hover Puck, Employments of hover puck, Innovation of hover puck, hockey puck was developed.  

Related links:  

Learn more about evolution on

brainly.com/question/727976

brainly.com/question/6953278

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(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

I= \frac{1}{2}mr^{2}

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

K= \frac{1}{2}I\omega ^{2}

Here, K is the rotational kinetic energy and \omega is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.356 m.

I= \frac{1}{2}(3.44)(0.356)^{2}

I= 0.218 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 0.218 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(0.218)(430) ^{2}

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

I= \frac{1}{2}mr^{2}

Put m= 3.44 kg and r= 0.775 m.

I= \frac{1}{2}(3.44)(0.775)^{2}

I= 1.03 kg m^{2}

Calculate the rotational kinetic energy of the smaller cylinder.

K= \frac{1}{2}I\omega ^{2}

Put I= 1.03 kg m^{2} and \omega = 430 rads^{-1}.

K= \frac{1}{2}(1.03)(430) ^{2}

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

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