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Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago

Potassium and chlorine are most likely to form what type of bond? ionic covalent metallic polar

fiasKO [112]

Answer:

The correct answer is the IONIC bond.

Explanation:

4 0

4 years ago

Read 2 more answers

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

4 years ago

Read 2 more answers

A long-distance runner is running at a constant speed of 5 m/s.

vivado [14]

Answer:

3.33 minutes (3 minutes and 20 seconds)

Explanation:

Speed of the runner = s = 5 m/s

We need to calculate how will it take for runner to complete 1 km. We have the speed, the distance and we need to find the time. Before performing any calculations, we must convert the values to same units.

Speed is in m/s and distance is in kilometers. So we have to either convert speed to km/s or distance into meters. In this case, converting distance into meters would be a convenient option.

1 kilo meters = 1000 meters

The distance, speed and time are related by the equation:

Distance = Speed x Time

So,

Time = Distance/Speed

Using the values, we get:

t = 1000/5

t = 200 seconds

This means, the runner can complete 1 kilometers in 200 seconds. Since, there are 60 seconds in a minute, we can convert this time to minutes, by dividing it by 60. i.e.

$200 \text{ sec} = \frac{200}{60} \text{ min} = 3.33 \text{ min}$