Physics, 24.06.2020 02:01 atlantahead7011

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.220 m from the center. He then mounts the tire in a vertical plane and notes that the radius vector to the spot is at an angle of 30.0Â° with the horizontal. Starting from rest, the tire is spun rapidly with a constant angular acceleration of 1.90 rad/s2. (Assume the spot's position is initially positive, and assume the angular acceleration is in the positive direction).A) What is the angular speed of the wheel after 1.30 s? B) What is the tangential speed of the spot after 1.30 s?
C) What is the magnitude of the total acceleration of the spot after 1.30 s?
D) What is the angular position of the spot after 1.30 s?

Answers: 2

Show answers

Answers

Answer from: kira3039

a) The angular speed of the wheel after 1.30 seconds is $2.47\,\frac{rad}{s}$ , b) The tangential speed of the spot after 1.30 seconds is $0.543\,\frac{m}{s}$ , c) The magnitude of the total acceleration of the spot after 1.30 seconds is $1.406\,\frac{m}{s^{2}}$ , d) The angular position of the spot is 2.130 radians (122.011°).

Explanation:

a) Given that tire accelerates at constant rate, final angular speed can be predicted by using the following formula:

$\omega = \omega_{o} + \alpha \cdot \Delta t$

Where:

$\omega$ - Final angular speed, measured in radians per second.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\Delta t$ - Time, measured in seconds.

Given that $\omega_{o} = 0\,\frac{rad}{s}$ (starts at rest), $\alpha = 1.90\,\frac{rad}{s^{2}}$ and $\Delta t = 1.30\,s$ , the final angular speed is:

$\omega = 0\,\frac{rad}{s} + \left(1.90\,\frac{rad}{s^{2}} \right) \cdot (1.30\,s)$

$\omega = 2.47\,\frac{rad}{s}$

The angular speed of the wheel after 1.30 seconds is $2.47\,\frac{rad}{s}$ .

b) The tangential speed of the spot is the product of the distance between the center of the wheel and spot. That is:

$v = r \cdot \omega$

Where r is the distance between the center of the wheel and spot. The tangential speed of the spot after 1.30 seconds is:

$v = (0.220\,m)\cdot \left(2.47\,\frac{rad}{s} \right)$

$v = 0.543\,\frac{m}{s}$

The tangential speed of the spot after 1.30 seconds is $0.543\,\frac{m}{s}$ .

c) The magnitude of the total acceleration of the spot is the magnitude of the vectorial sum of radial and tangential accelerations (both components are perpendicular to each other), which is determined by the Pythagorean theorem, that is:

$a = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where $a_{r}$ and $a_{t}$ are the radial and tangential accelerations.

$a = r\cdot \sqrt{\omega^{4} + \alpha^{2}}$

If $r = 0.220\,m$ , $\omega = 2.47\,\frac{rad}{s}$ and $\alpha = 1.90\,\frac{rad}{s^{2}}$ , then, the resultant acceleration is:

$a = (0.220\,m)\cdot \sqrt{\left(2.47\,\frac{rad}{s} \right)^{4}+\left(1.90\,\frac{rad}{s^{2}} \right)^{2}}$

$a \approx 1.406\,\frac{m}{s^{2}}$

The magnitude of the total acceleration of the spot after 1.30 seconds is $1.406\,\frac{m}{s^{2}}$ .

d) Let be 30° (0.524 radians) the initial angular position of the spot with respect to center. The final angular position is determined by the following equation of motion:

$\omega^{2} = \omega_{o}^{2} + 2\cdot \alpha \cdot (\theta - \theta_{o})$

Final angular position is therefore cleared:

$\theta - \theta_{o} = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}$

$\theta = \theta_{o} + \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}$

Given that $\theta_{o} = 0.524\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 2.47\,\frac{rad}{s}$ and $\alpha = 1.90\,\frac{rad}{s^{2}}$ , the angular position of the spot after 1.30 seconds is:

$\theta = 0.524\,rad +\frac{\left(2.47\,\frac{rad}{s} \right)^{2} - \left(0\,\frac{rad}{s}\right)^{2}}{2\cdot \left(1.90\,\frac{rad}{s^{2}} \right)}$

$\theta = 2.130\,rad$

$\theta = 122.011^{\circ}$

The angular position of the spot is 2.130 radians (122.011°).

Answer from: Quest

-first use of geothermal energy occurred more than 10,000 years ago in north america by american paleo-indians

-the first industrial use of geothermal energy began near pisa, italy in late 18th century

-in 1904, italian scientist piero ginori conti invented the first geothermal electric power plant in which steam was used to generate the power

-with the above experiment, the first geothermal plant in usa started in 1922 with a capacity of 250 kilowatts

-however, in 1946 first ground-source geothermal heat pump installed at commonwealth building in portland, oregon

-during the 1960’s, pacific gas and electric began operation of first large scale geothermal power plant in san francisco, producing 11 megawatts

-in 1973, when oil crisis began many countries began looking for renewable energy sources and by 1980’s geothermal heat pumps (ghp) started gaining popularity in order to reduce heating and cooling costs

-as effect of climate change started showing results, governments of various countries joined hands to fight against it, for which kyoto protocol was signed in japan in 1997, laid out emission targets for rich countries and required that they transfer funds and technology to developing countries, 184 countries have ratified it

hope this

Answer from: Quest

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

Another question on Physics

Physics, 21.06.2019 23:30

After a big snowfall, you take your favorite rocket-powered sled out to a wide field. the field is 195 m across, and you know that your sled accelerates at a rate of 3.65 m/s2 when the rocket is on. how much time will it take the sled to cross the field starting from rest, assuming the rocket is on the whole time?

Answers: 1

Answer

Physics, 22.06.2019 07:00

Aball has an initial velocity of 3 m/s. if there is no friction, what is the highest it could roll?

Answers: 1

Answer

Physics, 22.06.2019 14:00

Awater balloon launcher uses an elastic band with a spring constant of 115 n/m to use thee launcher you stretch thee band back with a balloon n it the water balloons have a mass of 1.3 kg a) if you stretch the band by 0.8 m, what elastic force will the band exert? b) with the band stretched 0.8 m, how much elastic potential energy is stored in the spring? c) if you release the band, it pushes forward. what is the kinetic energy of the balloon when it reaches the natural length of the band? d) what is the speed of the balloon when it reaches the natural length of the band?

Answers: 1

Answer

Physics, 22.06.2019 16:30

The sound level near a noisy air conditioner is 70 db. if two such units operate side by side, the sound level near the, would bea. 70dbb. 73 dbc. 105 dbd 140 db

Answers: 3

Answer

You know the right answer?

In testing an automobile tire for proper alignment, a technician marks a spot on the tire 0.220 m fr...

Questions