Learning Goal:

To understand work and power in rotational systems and to use the work-energy theorem to determine kinematics variables.

The variables used in standard linear mechanics (the study of objects that do not rotate) all have analogues in rotational mechanics (the study of objects that rotate). Here is a summary of variables used in kinematics and dynamics.

Linear variable Rotational variable
linear position: x angular position: θ
linear velocity: v angular velocity: ω
linear acceleration: a angular acceleration: α
linear inertia (mass): m moment of inertia: I
force: F torque: τ
The kinetic energy Krot associated with a rotating object is defined as

Krot=12Iω2,

where ω is measured in rad/s. Applying a torque can change the angular velocity of an object, and hence its kinetic energy. The torque is said to have done work on the object, just as a force applied over a distance can change the kinetic energy of an object in linear mechanics. In terms of rotational variables, the work W done by a constant torque τ is

W=τΔθ,

where Δθ describes the total angle, measured in radians, through which the object rotates while the torque is being applied.

Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2. Assume that the platform is initially at rest and the torque is applied for 12.0 rotations. Neglect friction.

Part A

How much work W does the motor do on the platform during this process?

Enter your answer in joules to four significant figures.

The questions are not complete so this is the complete questions

1. How much work W does the motor do on the platform during this process?

2. What is the angular velocity ωf of the platform at the end of this process?

3. What is the rotational kinetic energy, Ek, of the platform at the end of the process described above?

4. How long does it take for the motor to do the work done on the platform calculated in Part 1?

5. What is the average power delivered by the motor in the situation above?

6. . Note that the instantaneous power P delivered by the motor is directly proportional to ω, so P increases as the platform spins faster and faster. How does the instantaneous power P•f being delivered by the motor at the time t•f compare to the average power

P(average) calculated in Part e?

Explanation:

Given that,

The torque τ=25Nm

Moment of inertia I =50kgm²

The platform is initially at rest,

ω•i=0 rad/sec

Revolution the torque produce is 12

Then, θ=12 revolution

1 revolution=2πrad

So, θ=24πrad

1. Work done in a rotational motion is give as

W=τ•Δθ

Given that the τ=25Nm and the initial angular displacement is 0rad

The final angular displacement is 24πrad

Δθ =(θ2-θ1)

Δθ=24π-0

Δθ=24πrad

Then,

W=τ•Δθ

W=25(24π)

W=25×24π

W=1884.96J

To 4s.f, W=1885J

2. Final angular velocity ωf

Using the angular equation

ω•f²=ω•i²+2•α•Δθ

We need to get angular acceleration

The torque is given as

τ=I•α

Given that,

I is moment if inertia =50kgm²

τ=25Nm

α=τ/I

α= 25/50

α=0.5rad/s²

Now, using the angular acceleration

ω•f²=ω•i²+2•α•Δθ

ω•f²=0²+2×0.5×24π

ω•f²=0+75.398

ω•f²=75.398

ω•f=√75.398

ω•f=8.68 rad/sec.

3. We need to find rotational Kinetic energy and it is given as

K.E, = ½I•ω²

Given that, I=50kgm² and ω•f=8.68rad/sec

Then,

K.E, =½I•ω²

K.E, =½×50×8.68²

K.E, =1884.96J

To 4s.f,

K.E, =1885J

Which is the same as the work done by the motor.

4. Time taken to complete part 1,

Using the rotational equation

ω•f=ω•i+α•t

Since, ω•f=8.68 rad/sec and ω•i=0

And α=0.5rad/s²

Then,

ω•f=ω•i+α•t

8.68=0+0.5t

8.68=0.5t

Then, t=8.68/0.5

t=17.36secs

5. The average power of rotational motion is given as

P(average) =Workdone/timetaken

Since,

Work done =1884.96J

Time taken =17.36sec

P(average) =Workdone/timetaken

P(average)=1884.96/17.36

P(average)= 108.58Watts

To 4s.f

P(average)=108.6Watts

6. We need to find •, it is given as

• =τ•ωf

Given that, ω•f=8.68rad/sec, τ=25Nm

•=25×8.68

•=217Watts

Then, the ratio of • to P(average) is

Ratio = •/ P(average)

Ratio= 217/108.58

Ratio=1.9985

Then, the ratio is approximately 2

Ratio=2

think about the oscillation of the toy shown in the video. when the toy is at the highest point of its motion, its velocity is zero and its acceleration is negative.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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