Question

A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a time interval of 5 s, what is the angle in radians through which the wheel rotates? rad(c) At t=10 s the angular speed begins to increase at a rate of 1.6 rad/s/s. At t=15 s, what is the angular speed in radians per second? rad/s(d) Through what angle in radians did the wheel rotate during the time between t=10 s and t=15 s? rad(e) If the wheel rolls along the ground without slipping, the instantaneous velocity of the atoms of the object that are momentarily in contact with the ground is zero. This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R. During the time between t=10 s and t=15 s, how far did the center of the wheel move, in meters? m

Answer 1

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

Answer 2

The angular speed of this wheel with a radius of 36 cm is equal to 119.40 rad/s.

Given the following data:

• Radius = 36 cm to m = 0.36 m.

• Angular speed = 19 rev/s.

How to calculate angular speed in radians per second.

In order to calculate angular speed in radians per second (rad/s), we would have to convert the value in revolution per seconds (rev/s) to rad/s as follows:

$\omega = 19 \times 2\pi\\\\\omega = 19 \times 6.284$

Angular speed = 119.40 rad/s.

For part B:

• Time = 5 seconds.

Mathematically, the angular displacement through which the wheel rotates is given by:

∅ = ωt

∅ = 119.40 × 5

∅ = 597 rad.

For part C:

• Initial ime = 10 seconds.

• Angular acceleration = 1.6 rad/s²

• Final ime = 5 seconds.

Mathematically, the final angular speed is given by:

$\omega_f = \omega_i + \alpha \Delta t\\\\\omega_f =119.4 + 1.6(15-10)\\\\\omega_f =119.4 +1.6(5)\\\\\omega_f =119.4 + 8$

Final angular speed = 127.4 rad/s.

For part D:

• Initial ime = 10 seconds.

• Angular acceleration = 1.6 rad/s²

• Final ime = 5 seconds.

Mathematically, the angle through which the wheel rotates is given by:

$\theta = \omega_i \Delta t+\frac{1}{2} \alpha ( \Delta t)^2\\\\\theta =119.4(5)+\frac{1}{2} \times (1.6) \times 5^2\\\\\theta =597+20\\\\\theta =617\;rad$

For part E:

The average angular speed is calculated as follows:

$\omega =\frac{\omega_1 + \omega_2}{2} \\\\\omega =\frac{119.4 + 127.4}{2}\\\\\omega = 123.4\;rad/s$

Mathematically, the instantaneous speed of the contact point of the wheel is given by:

$v_{CM}=\omega R\\\\v_{CM}=123.4 \times 0.36\\\\v_{CM}=44.42\;m/s$

Now, we can determine the distance covered by the center of the wheel:

$Distance = v_{CM}t\\\\Distance = 44.42 \times (15-10)\\\\Distance = 44.42 \times 5$

Distance = 222.1 meters.

Read more on angular speed here: brainly.com/question/6860269