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IRISSAK [1]
4 years ago
5

If an object is not accelerating, then one knows for sure that it is_____.​

Physics
1 answer:
Sauron [17]4 years ago
6 0
The object is maintaining a constant velocity
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Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity fo
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Answer:

a) ω1 = 18rpm    ω2 = -18rpm

b) ω1 = 102rpm     ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm   This is the angular velocity of the robot

\omega = \frac{\omega r * D/2}{r_{wheel}}  where D is 30cm and rwheel is 5cm

\omega = \frac{6 * 30/2}{5}=18rpm  One velocity will be positive and the other will be negative:

ω1 = 18rpm    ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}

\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}

Replacing values, we get:

\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm

\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm

For part c, both wheels must have the same velocity:

\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min

\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm

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A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If
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A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts from rest and is then brought in contact with a spinning rubber wheel, we observe that the disk gradually begins to rotate too. If after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 4.0 rad/s, find the average angular acceleration that the disk experiences. (Assume the positive direction is in the initial direction of rotation of the disk. Indicate the direction with the sign of your answer.)

Assume after 35 s of contact with this spinning rubber wheel, the disk has an angular velocity of 11.0 rad/s.

Answer:

385 rad

Explanation:

The expression for the angular acceleration of a disk that is in contact with a spinning wheel can be given as:

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

where \delta\omega = \omega_f - \omega_i

\alpha = \frac{ \omega_f-\omega_i}{\delta t}

\alpha = \frac{ 4.0 rad/s-0 rad/s}{35}

\alpha =0.14 rad/s^2

Angular displacement of a disk can be calculated by using the formula:

\theta = \omega t

substituting 11.0 rad/s for \omega and t = 35 s ; we have:

\theta = 11.0 rad/s * 35 s

\theta = 385 rad

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