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ad-work [718]
3 years ago
15

What’s is the relationship between energy and motion ?

Physics
1 answer:
LekaFEV [45]3 years ago
5 0
Motion energy is the sum of potential and kinetic energy in an object that is used to do work.
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A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989

4 0
3 years ago
ANSWER QUICK 30 POINTS
Lelu [443]
What do you need help on
4 0
3 years ago
What is a an reason why a object gets extinct
adoni [48]

Answer:

An object may become extinct due to the lack of quantity of the object and the object no longer exists so it is now extinct.

7 0
3 years ago
To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build
Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3>y=v_{o} *t+\frac{1}{2} *g*t^{2}</h3><h3>y: The vertical distance the ball moves at time t  </h3><h3>v_{o}i: Initial speed </h3><h3>g=Gravity acceleration=9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )</h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s  is equal to height of Sears Tower (h).  

Too we know that the ball is released from rest, then,v_{0}=0

Height of Sears Tower calculation:

We replace  in the equation 1 the data following;

y=h

v_{o} =0

g=32,1\frac{ft}{s^{2} }

t= 9,5s

h=0*9.5+\frac{1}{2} *32.1*9.5^{2}

h=1448.5 ft

Answer: The height of Sears Tower is 1448.5 ft

6 0
3 years ago
A 0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and b
Ksenya-84 [330]
We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt 
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N 


To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N 

We can see that weight is much smaller than the applied force so it's influence in negligible.
3 0
3 years ago
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