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2 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Physics

1 answer:

Ede4ka [16]2 years ago

6 0

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

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The mechanical energy isn't conserved. Some energy is lost to friction.

Option A.

<h3>Explanation:</h3>

The mechanical energy is defined as the energy of a body which it achieves by virtue of its position and velocity. The mechanical energy are of two types - potential energy and kinetic energy. The potential energy is the energy of the body which it achieves by means of its relative position and is directly proportional to the height of the body from its relative plane. Whereas the kinetic energy of the body is achieved by virtue of its velocity and is directly proportional to the square of velocity of the body.

As the mountaineer is skiing down the slope of a mountain, the potential energy of the person is gradually changing into his kinetic energy. Had it been in an ideal situation, the potential energy lost would have been just equal to the kinetic energy gained by the person. But there's friction which opposes the speed of the body and reduces the velocity. Thus the kinetic energy will be lost to some extent and the energy won't be conserved.

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3 years ago

the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the tab

JulsSmile [24]

The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

Mass of the first block, m₁ = 4.0 kg
Mass of the second block, m₂ = 2.0 kg

The net force on the system of the two blocks is calculated as follows;

$m_2 g - T = m_1 a$

where;

T is the tension in the connecting string due weight of the first block

$m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2$

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

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2 years ago

A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw

navik [9.2K]

Answer: 75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

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momentum of the Lion and Gazelle = 200,900.7 x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 = 200,900.7 x V

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V = 75,242.9 m/s

7 0

2 years ago

Read 2 more answers

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.

Alexxandr [17]

Answer:

Explanation:

Given that,

B(t) = B0 cos(ωt) • k

Radius r = a

Inner radius r' = a/2 and resistance R.

Current in the loop as a function of time I(t) =?

Magnetic flux is given as

Φ = BA

And the Area is given as

A = πr², where r = a/2

A = πa²/4

Then,

Φ = ¼ Bπa²

Φ(t) = ¼πa²Bo•Cos(ωt)

Then, the EMF is given as

ε(t) = -dΦ/dt

ε(t) = -¼πa²Bo • -ωSin(ωt)

ε(t) = ¼ωπa²Bo•Sin(ωt)

From ohms law,

ε = iR

Then, i = ε/R

I(t) = ¼ωπa²Bo•Sin(ωt) /R

This is the current induced in the loop.

Check attachment for better understanding

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3 years ago

A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr

quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = $\frac{n}{2L} x \sqrt{\frac{T}{μ}}$

f = $\frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }$

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

8 0

3 years ago