Answer:
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
