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goldfiish [28.3K]
3 years ago
8

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

s subjected to a torque M = (t2+2) N⋅m, where t is in seconds. Each ball has a speed v = 2 m/s when t = 0.
Determine the speed of each ball when t = 3 s.

Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
Digiron [165]3 years ago
7 0

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

L = L_i_n_i + L_3_s

L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt

L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

Moment if inertia is

I = 2ml^2

I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

Angular speed

ω = L/I

= 38 / 5.75\\\\=6.61

The speed of each ball is

V = ωL

= 6.61\times0.5\\\\= 3.31m/s

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A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has
irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration (a) consist in two components, <em>radial</em> (a_{r}) and <em>tangential</em> (a_{t}), in meters per square second:

a_{r} = \omega^{2}\cdot r (1)

a_{t} = \alpha \cdot r (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

a = \sqrt{a_{r}^{2}+a_{t}^{2}} (3)

Where:

r - Radius of the wheel, in meters.

\omega - Angular speed, in radians per second.

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

Given that wheel accelerates uniformly, we use the following kinematic equation:

\omega = \omega_{o}+ \alpha\cdot t (4)

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t - Time, in seconds.

If we know that r = 0.1\,m, \alpha = 2\,\frac{rad}{s^{2}}, \omega_{o} = 0\,\frac{rad}{s} and t = 0.60\,s, then the total linear acceleration is:

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

\omega = 1.2\,\frac{rad}{s}

a_{r} = \omega^{2}\cdot r

a_{r} = 0.144\,\frac{m}{s^{2}}

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3 0
3 years ago
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s
IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

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next we will express the distance between the earth and the sun

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a = 149.668 * 10^8

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input the given values into equation 1 above

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next calculate the Earths velocity of approach towards the sun using this equation

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Rc = 149.626 * 10^9 m

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(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

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