Solving Angular Rotation Homework: Question b, c & d

[ataglance05]

Homework Statement

A 1kg engine rotates at 1 rad/sec about a point on a 3meter wire. The engine produces an acceleration of 1 rad/sec^2 and is fird for 2 minutes.

a) What is the ω at the end of 2 minutes?
b) what's the tension in the wire at the end of 2 min.?
c) how far will the engine travel in meters while it's accelerating?
d) what's the Kinetic Energy of the engine after it shuts down?

Homework Equations

a)ω=radius/Time or ω= θ/Time
B) ?
c) S=rθ
θ= ωi(T)+1/2αT^2
d) KE=1/2(Ι)(angluar acceleration)(T)^2
Ι=1/2mr^2

The Attempt at a Solution

a) ω=radius/Time
ω=3/120= 0.25 rad/sec
b)?
c) ?
d) Ι=1/2mr^2
Ι=1/2(1)(3)^2= 3

KE=1/2(Ι)(angluar acceleration)(T)^2
KE=1/2(3)(1)(120)^2
KE=21600 J

need help with b&c and with corroborating a&d!
thanks!

 

[andrevdh]

a) The engine is experiencing a constant angular acceleration, $$\alpha$$, while it is fired. The constant angular acceleration equations look similar to the constant linear acceleration equations:

$$\omega = \omega _i + \alpha t$$

b) Use the final angular speed to calculate the centripetal force that the engine is experiencing, this will be the tension in the wire.

c) your equations should do the trick

d) I think the kinetic energy is given by

$$EK = \frac{1}{2}I \omega ^2$$

 

[m2003]

b) To find the tension in the wire, we can use the formula T=mrα, where T is the tension, m is the mass of the engine (1kg), r is the radius of rotation (3m), and α is the angular acceleration (1 rad/sec^2). Plugging in these values, we get T=1(3)(1)=3N. So, the tension in the wire at the end of 2 minutes is 3 Newtons.

c) To find the distance traveled by the engine while it is accelerating, we can use the formula S=rθ, where S is the distance, r is the radius of rotation (3m), and θ is the angular displacement. To find θ, we can use the formula θ= ωi(T)+1/2αT^2, where ωi is the initial angular velocity (1 rad/sec), T is the time (2 minutes = 120 seconds), and α is the angular acceleration (1 rad/sec^2). Plugging in these values, we get θ= (1)(120)+1/2(1)(120)^2= 120+7200= 7320 rad. Now, plugging this value into the first formula, we get S= (3)(7320)= 21960 meters. So, the engine will travel 21960 meters while it is accelerating.

d) To find the kinetic energy of the engine after it shuts down, we can use the formula KE=1/2(Ι)(ω)^2, where Ι is the moment of inertia (1/2mr^2), and ω is the angular velocity (0.25 rad/sec). Plugging in these values, we get KE=1/2(1/2)(1)(3)^2(0.25)^2= 0.09375 J. So, the kinetic energy of the engine after it shuts down is 0.09375 Joules.