Ask question

Ask question

All categories

4 years ago

5

What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg⋅

m2, and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?

1 answer:

mart [117]4 years ago

7 0

The complete question is;

To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0º . What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg.m², and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?

Answer:

α = 42.76 rad/s²

Explanation:

We are given;

Mass; m = 2 kg

Distance; r = 24cm = 0.24m

Moment of inertia of the forearm; I = 0.24 kg.m²

Muscle Force; F = 759N

Perpendicular distance to lever arm; r' = 2cm = 0.02m

Angle at which she flexes arm; θ =60°

Let's assume that the arm starts extended vertically downwards. and we take the elbow joint as the point about which we calculate the torque.

Thus, we know that torque is given by the formula ;

τ = Force x Perpendicular distance

Thus, the toque exerted by force in the muscle is;

τ_muscle = 759 x 0.02 = 15.18 N.m

Also, torque exerted by the lifted weight is given as 0 because perpendicular distance is zero.

Thus, the net torque on the lower arm is;

τ_net = τ_muscle - τ_weight

τ_net = 15.18 - 0 = 15.18 N.m

Now, let's calculate moment of inertia of lifted weight.

The moment of inertia is given by;

I = mr²

Thus, moment of Inertia of lifted weight is; I_weight = 2 x 0.24² = 0.115 kg.m²

Thus,total moment of inertia is;

I_total = I_arm + I_weight

Thus, I_total = 0.24 + 0.115 = 0.355 kg.m²

Now, we can calculate the angular acceleration.

It's gotten from the equation;

τ_net = I_total•α

Where α is angular acceleration.

Thus making α the subject, we have ; α = τ_net/I_total

α = 15.18/0.355 = 42.76 rad/s²

You might be interested in

Process of diffusion using particle theory

irinina [24]

Answer:

spreading a substance from a region of high concentration to a region of low concentration.

Explanation:

Diffusion is a process of spreading a substance from a region of high concentration to a region of low concentration. It occurs when the particles of the substance move through the space between the particles of another substance.

7 0

2 years ago

True or false day and night are caused by earths rotation

olga2289 [7]

True because it is day in one part and night in the other

7 0

4 years ago

Read 2 more answers

Explain the difference between rotation and revolution. Then use these terms to explain how earth

natima [27]

Answer:

The difference between rotations and revolutions is , when an object turns around an internal axis (like the Earth turns around its axis) it is called a rotation. When an object circles an external axis (like the Earth circles the sun) it is called a revolution.

Explanation:

While rotation means spinning around its own axis, revolution means to move around another object. Taking the example of the Earth, which rotates 366 times to complete one revolution around the Sun.

8 0

4 years ago

A) One Strategy in a snowball fight the snowball at a hangover level ground. While your opponent is watching this first snowfall

Alexandra [31]

Answers:

a) $\theta_{2}=38\°$

b) $t=0.495 s$

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the snowball has two components: <u>x-component</u> and <u>y-component</u>. Being their main equations as follows for both snowballs:

<h3><u>Snowball 1:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{1} t_{1}$ (1)

Where:

$V_{o}=14.1 m/s$ is the initial speed of snowball 1 (and snowball 2, as well)

$\theta_{1}=52\°$ is the angle for snowball 1

$t_{1}$ is the time since the snowball 1 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the snowball 1 (assuming that both people are only on the x axis of the frame of reference, therefore the value of the position in the y-component is zero.)

$y=0$ is the final height of the snowball 1

$g=-9.8m/s^{2}$ is the acceleration due gravity (always directed downwards)

<h3><u>Snowball 2:</u></h3>

<u>x-component: </u>

$x=V_{o}cos\theta_{2} t_{2}$ (3)

Where:

$\theta_{2}$ is the angle for snowball 2

$t_{2}$ is the time since the snowball 2 is thrown until it hits the opponent

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta_{2} t_{2}+\frac{gt_{2}^{2}}{2}$ (4)

Having this clear, let's begin with the answers:

<h2>a) Angle for snowball 2</h2>

Firstly, we have to isolate $t_{1}$ from (2):

$0=0+V_{o}sin\theta_{1} t_{1}+\frac{gt_{1}^{2}}{2}$ (5)

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$ (6)

Substituting (6) in (1):

$x=V_{o}cos\theta_{1}(-\frac{2V_{o}sin\theta_{1}}{g})$ (7)

Rewritting (7) and knowing $sin(2\theta)=sen\theta cos\theta$ :

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{1})$ (8)

$x=-\frac{(14.1 m/s)^{2}}{-9.8 m/s^{2}} sin(2(52\°))$ (9)

$x=19.684 m$ (10) This is the point at which snowball 1 hits and snowball 2 should hit, too.

With this in mind, we have to isolate $t_{2}$ from (4) and substitute it on (3):

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$ (11)

$x=V_{o}cos\theta_{2} (-\frac{2V_{o}sin\theta_{2}}{g})$ (12)

Rewritting (12):

$x=-\frac{V_{o}^{2}}{g} sin(2\theta_{2})$ (13)

Finding $\theta_{2}$ :

$2\theta_{2}=sin^{-1}(\frac{-xg}{V_{o}^{2}})$ (14)

$2\theta_{2}=75.99\°$

$\theta_{2}=37.99\° \approx 38\°$ (15) This is the second angle at which snowball 2 must be thrown. Note this angle is lower than the first angle $(\theta_{2} < \theta_{1})$ .

<h2>b) Time difference between both snowballs</h2>

Now we will find the value of $t_{1}$ and $t_{2}$ from (6) and (11), respectively:

$t_{1}=-\frac{2V_{o}sin\theta_{1}}{g}$

$t_{1}=-\frac{2(14.1 m/s)sin(52\°)}{-9.8m/s^{2}}$ (16)

$t_{1}=2.267 s$ (17)

$t_{2}=-\frac{2V_{o}sin\theta_{2}}{g}$

$t_{2}=-\frac{2(14.1 m/s)sin(38\°)}{-9.8m/s^{2}}$ (18)

$t_{2}=1.771 s$ (19)

Since snowball 1 was thrown before snowball 2, we have:

$t_{1}-t=t_{2}$ (20)

Finding the time difference $t$ between both:

$t=t_{1}-t_{2}$ (21)

$t=2.267 s - 1.771 s$

Finally:

$t=0.495 s$

4 0

4 years ago

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690

Gwar [14]

Answer:

61.4 s

Explanation:

The distance d₁ traveled by the asteroid:

$d_1=\frac{1}{2}a_1t^2$

The distance d₂ traveled by the space ship:

$d_2=\frac{1}{2}a_2t^2$

The total distance d:

$d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2$

Solving for time t:

$t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}$

3 0

3 years ago

Read 2 more answers