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joja [24]
3 years ago
6

A wheel rotates clockwise 6 times per second. What will be its angular displacement after 7 seconds? Answer should be rounded to

2 decimal places
Physics
2 answers:
Yuri [45]3 years ago
8 0

(6 rotations/sec) x (7 sec) = 42 rots

Each rotation is 360 degrees or 2π radians.

42 rotations = 15,120 degrees

or

84π radians .

solong [7]3 years ago
5 0

Answer:

The frequency of the wheel is the number of revolutions per second:

f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz  

And now we can calculate the angular speed, which is given by:

\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.

Explanation:

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A disk rotates at constant angular acceleration, from angular position θ1 = 16.0 rad to angular position θ2 = 76.0 rad in 5.30 s
Oliga [24]

Answer:

(a) the angular velocity at θ1 is 11.64 rad/s

(b) the angular acceleration is 0.12 rad/s^{2}

(c) the angular position was the disk initially at rest is - 428.27 rad

Explanation:

Given information :

θ1 = 16 rad

θ2 = 76 rad

ω2 = 11 rad/s

t = 5.3 s

(a) The angular velocity at θ1

First, we use the angular motion equation for constant acceleration

Δθ = (ω1+ω2)t/2

θ2 - θ1 = (ω1+ω2)t/2

ω1 + ω2 = 2 (θ2 - θ1) / t

ω1 = (2 (θ2 - θ1) / t ) - ω2

     = (2 (76-16) / 5.3) - 11

     = 11.64 rad/s

(b) the angular acceleration

ω2 = ω1 + α t

α t = ω2 - ω1

α = (ω2 - ω1)/t

  = (11.64 - 11) / 5.3

  = 0.12 rad/s^{2}

(c) the angular position was the disk initially at rest, θ0

at rest ω0 = 0

ω2^2 = ω01  t + 2 α Δθ

2 α Δθ = ω2^2

θ2 - θ0 = ω2^2  /  2 α

θ0 = θ2 -  (ω2^2) / 2 α

  = 76 - (11^{2}/ 2 x 0.12

  = 76 - 504.16

  = - 428.27 rad

4 0
3 years ago
A bicycle wheel rotates at a constant 25 rev/min. What is true about its angular acceleration?
Nostrana [21]

Answer:

The angular acceleration is zero

Explanation:

When an object is in rotational motion, it has a certain angular velocity, which is the rate of displacement of its angular position.

This angular velocity can change or remain constant - this is given by the angular acceleration, which is:

\alpha =\frac{\Delta \omega}{\Delta t}

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\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

Therefore, the angular acceleration is the rate of change of angular velocity.

In this problem, the bicycle rotates at a constant angular velocity of

\omega=25 rev/min

This means that the change in angular velocity is zero:

\Delta \omega=0

And so, that the angular acceleration is zero:

\alpha=0

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