Speed and accleration of a turntable

[clope023]

Homework Statement

An electric turntable 0.740m in diameter is rotating about a fixed axis with an initial angular velocity of 0.290rad/s . The angular acceleration is 0.900rad/s^2 .

a) Compute the angular velocity after a time of 0.194s. - .465rev/s

b) Through how many revolutions has the blade turned in this time interval? - .073rev

c) What is the tangential speed of a point on the tip of the blade at time t = 0.194s ?

d) What is the magnitude of the resultant acceleration of a point on the tip of the blade at time = 0.194 ?

Homework Equations

v = r$$\omega$$

arad = $$\omega$$^2r

atan = r$$\alpha$$

The Attempt at a Solution

for part c)

v = (.740/2)(.465) = .172m/s wrong

for part d)

arad = (.465)^2(.740/2) = .0800m/s^2

atan = (.740/2)(.900) = .333m/s^2

a = (.0800^2 + .333^2)^(1/2) = .342m/s^2 wrong

I'm not sure what I'm doing wrong, any help is appreciated.

 

[bgask001]

For part C it seems as if you need to do is convert the revs to meters. To do this you need to know the circumference (which would be the distance "unrolled" after 1 revolution)

 

[Moetasim]

Hello! I can help you with your questions about the speed and acceleration of a turntable. It seems like you have attempted to use the correct equations, but there may be some errors in your calculations. Let's go through each part together to find the correct answers.

a) To compute the angular velocity after a time of 0.194s, we can use the equation:

\omega_f = \omega_i + \alpha t

Where \omega_f is the final angular velocity, \omega_i is the initial angular velocity, \alpha is the angular acceleration, and t is the time. Plugging in the given values, we get:

\omega_f = 0.290 + (0.900)(0.194) = 0.443 rad/s

b) To find the number of revolutions, we can use the equation:

\theta = \omega_it + \frac{1}{2}\alpha t^2

Where \theta is the angle of rotation, \omega_i is the initial angular velocity, \alpha is the angular acceleration, and t is the time. Plugging in the given values, we get:

\theta = (0.290)(0.194) + \frac{1}{2}(0.900)(0.194)^2 = 0.073 rev

c) To find the tangential speed at time t = 0.194s, we can use the equation:

v = r\omega

Where v is the tangential speed, r is the radius, and \omega is the angular velocity. Plugging in the given values, we get:

v = (0.740/2)(0.443) = 0.163 m/s

d) To find the magnitude of the resultant acceleration, we can use the Pythagorean theorem:

a = \sqrt{a_{rad}^2 + a_{tan}^2}

Where a is the magnitude of the resultant acceleration, a_{rad} is the radial acceleration, and a_{tan} is the tangential acceleration. Plugging in the given values, we get:

a = \sqrt{(0.900)^2(0.740/2)^2 + (0.740/2)(0.443)^2} = 0.393 m/s^2

I hope this helps clarify any confusion and provides you with the