what is the torque exerted on it? please PLEASE help. due tomorrow!!?
when the play butotn is pressed , a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. if the CD has a radius of 6.0 cm and a mass of 17 g. what is the torque exerted on it?
i know the answer is 0.0018Nm but i need to know how to work it out. so if u could PLEASE show me like all the steps u did that would be cool
- Dr DLv 71 decade agoFavorite Answer
Torque = I*α
I = 1/2 * m * r^2 = 1/2 * 0.017 * 0.06^2 kg m^2
To find α we can adjust Newton's equaiton of motion to circular motion
ω^2 = 0 + 2*α * θ
(450*2π/60)^2 = 2 * α * 3*2π
α = 58.9 rad/sec^2
Torque = 1.80 x 10^-3 Nm
- Anonymous1 decade ago
I think I can help you.
The torque required can be computed using the following equation.
tau = alpha*Iz
where tau is the torque
alpha is the acceleration
Iz is the moment of inertia for a cylinder
Using reference 1, we see that
Iz = 1/2*m*r^2 = 3.06*10^-5 *kg*m^2
Where m = 18 grams
r = 6 cm
Alpha is the term that requires a bit of thought. I will work by analogy. If we were working with linear variables, we can say that
vf^2 - vi^2 = 2*a*d
Where vf is the final velocity of an object
vi is the initial velocity of an object
a is its acceleration
d is the distance it moves.
A similar relationship exists for angular velocity.
wf^2 - wi^2= 2*alpha*theta
Where wf is the final angular velocity (450*2*pi/sec)
wi is the initial angular velocity (0 rad/sec)
theta is the angle the object moves (3*2*pi)
Solving for alpha, I get alpha = 58.905 rad/sec^2.
Substituting into the initial equation, I get tau = 0.0018 N*mSource(s): 1. Moment of inertia for a cylinder http://en.wikipedia.org/wiki/List_of_moments_of_in...
- gp4rtsLv 71 decade ago
The average angular velocity is 450/2 = 225 rev/min = 3.75 rev/sec; The no of rotations at this average is 225*t. Therefore the time it takes to make these revs is 3 = 3.75*t, or t = 0.08 sec. The angular acceleration is 7.5 (rev/sec)/.8 (sec) (It takes 0.8 sec to go from 0 to 7.5 rps). This is a = 9.375 rev/sec^2
The formula for acceleration of a rotating body is
T = I*a, were I is the moment of inertia. For a thin disc, the moment of inertia is I = M*r^2 / 2. Convert to std units M = 0.017 kg, and r = 0.06 m gives I = 3.06*10^-5 kg*m^2
Convert angular acceleration to rad / sec^2. I rev = 2*π radians. This gives 9.375*2*π = 58.905 rad/sec^2
Now compute the torque T = 58.905 * 3.06*10^-5 = 0.0018 N*m
- EdwardLv 71 decade ago
I-moment of itertia
A- (alpha) angular acceleration
aT- tangential acceleration
R - radius
at=( 2pi R rpm/60) 3 rev =( 2x 3.14....x 06 x 450/ 60)/ 3
at= .94 m/sec^2
A= aT/ R= .94 m/sec^2 /(.06 m)=15.7rad/sec^2
I=0.5 m R^2= 0.5 x 0.017 x (0.6)^2=0.0000306kg m^2
T= I A=0.0000306kg m^2 x 15.7 rad/sec^2
T= 480 E-6 mNSource(s): 1. http://www.physics.uoguelph.ca/tutorials/torque/Q.... 2. http://en.wikipedia.org/wiki/Angular_acceleration
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- 1 decade ago
take acceleration =velo/sec=450/60 re/sxs
torque=Moment of Inertia (angular acceleration )
M.I for disc with refeence to tangent in the plane =moment of inertia with reference to doameter=MR(R)/4
torque= (Mr x r)/4 X(a/r) here a/r is the angular acceleration
substitute the values you will get the answer
convert radius into meter and mass into kilogram
bye- enjoy :Physics lecturerSource(s): Yahoo
- Hootie JLv 51 decade ago