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Snowcat [4.5K]
3 years ago
12

A stick is resting on a concrete step with 2/5 of its length hanging over the edge. A single ladybug lands on the end of the sti

ck hanging over the edge, and the stick begins to tip. A moment later, a second, identical ladybug lands on the other end of the stick, which results in the stick coming momentarily to rest 41.3° from the horizontal. If the mass of each bug is 3.43 times the mass of the stick and the stick is 18.7 cm long, what is the magnitude of the angular acceleration of the stick at the instant?
Physics
1 answer:
Trava [24]3 years ago
3 0
The moment the stick comes to rest at θ=62.1° from horizontal. 

<span>Angular acceleration = (net torque) / (moment of inertia) </span>
<span>α = τ/I </span>

<span>We have to add up the torques due to the bugs and the stick; and add up the moments of inertia due to all three also. </span>

<span>Let L be the stick's length and let m be the stick's mass (so "2.75m" is each bug's mass). And let's say the "lower" ladybug is on the left. Then the lower ladybug exerts this much torque: </span>

<span>τ_lowerbug = −(2/5)L(2.75mg)cosθ (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction). </span>

<span>The upper ladybug exerts this much torque: </span>

<span>τ_upperbug = +(3/5)L(2.75mg)cosθ </span>

<span>The weight of the stick can be assumed to act through its center, which is 1/10 of the way from the fulcrum. So the stick exerts this much torque: </span>

<span>τ_stick = +(1/10)L(mg)cosθ </span>

<span>The net torque is thus: </span>

<span>τ_net = τ_lowerbug + τ_upperbug + τ_stick </span>
<span>= −(2/5)L(2.75mg)cosθ + (3/5)L(2.75mg)cosθ + (1/10)L(mg)cosθ </span>
<span>= (2.75(3/5−2/5)+1/10)(mgL)cosθ </span>

<span>Now for the moments of inertia. The bugs can be considered point masses of "2.75m" each. So for each of them you can use the simple formula: I=mass×R²: </span>

<span>I_lowerbug = (2.75m)((2/5)L)² = (2.75m)(4/25)L² </span>
<span>I_upperbug = (2.75m)((3/5)L)² = (2.75m)(9/25)L² </span>

<span>For the stick, we can use the parallel axis theorem. This says, when rotating something about an axis offset a distance "R" from its center of mass, the moment of inertia is: </span>

<span>I = I_cm + mR² </span>

<span>We know that for a stick about its center of mass, I_cm is (1/12)mL² (see many sources). And in this problem we know that it's offset by R=(1/10)L. So: </span>

<span>I_stick = (1/12)mL² + m((1/10)L)² </span>
<span>= (1/12)mL² + (1/100)mL² </span>
<span>= (7/75)mL² </span>

<span>So the total moment of inertia is: </span>

<span>I_total = I_lowerbug + I_upperbug + I_stick </span>
<span>= (2.75m)(4/25)L² + (2.75m)(9/25)L² + (7/75)mL² </span>
<span>= (2.75(4/25+9/25)+7/75)mL² </span>

<span>So that means the angular acceleration is: </span>

<span>α = τ_net/I_total </span>
<span>= ((2.75(3/5−2/5)+1/10)(mgL)cosθ)/((2.75(4... </span>

<span>The "m" cancels out. You're given "L" and "θ" and you know "g", so do the math (and don't forget to use consistent units).</span>
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Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0
3 years ago
you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0
3 years ago
A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express
andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

      E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

     E = mcΔT

     725 = 0.055 x 921.096 x (T - 27.5)

     T - 27.5 = 14.31

     T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

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3 years ago
Which force requires contact?
Minchanka [31]

Answer:

A

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I think it’s 15cm
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