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STALIN [3.7K]
3 years ago
6

If an ant is starting from 0 radians, and travels all the way to 4.5 radians in 18 seconds, what is angular velocity? If the sam

e ant then slows down back to 0 rad/s in 3 seconds what is its angular acceleration?
Physics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

  • Angular velocity is 0.25 rad/s
  • Angular acceleration is 0.017 rad/s²

Explanation:

Given;

initial angular displacement, θ₁ = 0 radians

final angular displacement, θ₂ = 4.5 radians

Angular velocity is calculated as;

\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18}  \\\\\omega = 0.25 \ rad/s

Angular acceleration is calculated as;

\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}

where;

\omega_f is the final angular velocity = 0 rad/s

\omega _i is the initial angular velocity = 0.25 rad/s

t₂ is the final time of the motion = 3 seconds

t₁ is the initial time of the motion = 18 seconds

\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2

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A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl
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Answer:

<em>10.90km</em>

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

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dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

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6 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

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