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Lilit [14]
3 years ago
12

Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father

exerts a force on the merry-go-round perpendicular to its radius to achieve an angular acceleration of 4.44 rad/s2. (a) how long (in s) does it take the father to give the merry-go-round an angular velocity of 1.43 rad/s? (assume the merry-go-round is initially at rest.)

Physics
2 answers:
Dmitriy789 [7]3 years ago
8 0

It takes 0.322 s to give the merry-go-round an angular velocity of 1.43 rad/s

\texttt{ }

<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

moment of inertia = I = 84.4 kg.m²

initial angular velocity = ωo = 0 rad/s

angular acceleration = α = 4.44 rad/s²

final angular velocity = ω = 1.43 rad/s

<u>Asked:</u>

time taken = t = ?

<u>Solution:</u>

\omega = \omega_o + \alpha t

1.43 = 0 + 4.44t

1.43 = 4.44t

t = 1.43 \div 4.44

t = 0.322 \texttt{ s}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

-Dominant- [34]3 years ago
6 0
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
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The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from
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Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) V_{max} the car has is:

V_{max}=203 mph=90.74 m/s taking into account 1 mile=1609.34 m

And the car's average acceleration a_{ave} is:

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Since:

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Where:

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Finding this time:

\Delta t=\frac{V_{f}-V_{o}}{a_{ave}} (2)

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t_{TOTAL}=t_{1}+ t_{2} (15)

t_{TOTAL}=101.95 s + 169.45 s (16)

t_{TOTAL}=271.4 s s

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