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Elina [12.6K]
2 years ago
11

A boy is swinging a toy on a piece of string in a vertical circle. The toy has a mass of 150 g and the radius of the circle is 0

.8 m. a) He swings the toy with a linear velocity of 2 m/s. Will the toy move in a circle? Explain your answer. b) Another boy swings the toy with a linear velocity of 3.5 m/s. Work out the tension in the string at the top of the circle, at the bottom of the circle and halfway between the top and the bottom of the circle.​
Physics
1 answer:
Maru [420]2 years ago
3 0
At the top of the circular motion, both weight and tension provides for centripetal force.

By Newton’s Second Law,
Fnet = ma
mg + T = mv^2/r (since a = v^2/r and weight = mg)

For toy to continue moving in circle at the top,

T > 0
mv^2/r - mg > 0
v >root (gr)

Hence, minimum speed toy must have is 2.80 m/s. Since linear velocity is lower than the minimum linear velocity, the toy will not move in circular motion.

b) Tension at top = mv^2/r - mg
= (0.15)(3.5)^2/0.8 - (0.15)(9.81)
= 0.825 N

Tension at bottom = mv^2/r + mg
= (0.15)(3.5)^2/0.8 + (0.15)(9.81)
= 3.77 N

In the middle, only Tension provides for centripetal force. Hence,
Tension = mv^2/r
= (0.15)(3.5)^2/0.8
= 2.30 N
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A uniform cylinder of radius 25 cm and mass 27 kg is mounted so as to rotate freely about a horizontal axis that is parallel to
Alisiya [41]

Answer:

Explanation:

                                                     STEP 1

<u>Given</u>

Radius of cylinder = r = 25cm, 2.5m

mass = 27kg

cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder

height = 0.6m

<u>part 1</u>

The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by

<em>I = Icm + mh²</em>

<em>∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20  x  (0.6)²</em>

<em>I=13.09kg.m²</em>

where

<em>I</em>cm is the rotational inertia of the cylinder about its central axis

m is the mass of the cylinder

h is the distance between the axis of the rotation and the central axis of the cylinder

r is the radius of the cylinder

<em>                                        </em><em> I=13.09kg.m²</em>

<em>part2</em>

<em>from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero</em>

<em>Δk + Δu = 0</em>

<em>1/2 </em>I(w²-w²) = Ui-Uf

1/2 x 13.09w² = mgh

∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5

w=18.09rad/s

5 0
3 years ago
A turtle accelerates from a stop at 3m/s/s to the South for 8s. What is the turtle’s final velocity? Show your work and include
andrey2020 [161]

He's accelerating at 3 m/s² .  That means his speed is increasing by 3 m/s every second.At the end of 8 seconds, his speed is (8 x 3 m/s)  =  24 m/s .He's been moving south for the whole 8 seconds.So at the end of that time, his velocity is                          24 m/s south .

5 0
2 years ago
A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1
lidiya [134]

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as  

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and  

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people  

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum  

2000 ×  1 = (2000 + 3000 ) ω

solve we get now  

ω = 0.4 rad/s  

5 0
3 years ago
What usually results when an organism fails to maintain homeostasis?
dedylja [7]
The organism may become ill or die
3 0
3 years ago
Read 2 more answers
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
3 years ago
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