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zaharov [31]
3 years ago
13

The 100-kg homogeneous cylindrical disk is at rest when the force is applied to a cord wrapped around it, causing the disk to ro

ll. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution.
Physics
1 answer:
Ivahew [28]3 years ago
3 0

Complete question:

The 100-kg homogeneous cylindrical disk is at rest when the force F =500N is applied to a cord wrapped around it, causing the disk to roll. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution (radius of the disk = 300mm).

Answer:

The angular velocity of the disk when it has turned one revolution is 16.712 rad/s

Explanation:

From the principle of work and energy;

U = E₂ - E₁, since the disk is initially at rest, T₁ = 0

U = E₂

Work done, U = product of force and perpendicular distance

U = F × d

As the cord winds, force act through the cord at a distance of 2d

U = F × 2d

Distance of one complete revolution = 2πR = 2π(0.3) = 0.6π

U = 500 × 2(0.6π) = 1885.2 J

Kinetic energy E₂ = \frac{1}{2}I \omega^2 +  \frac{1}{2}m v^2

E_2 = \frac{1}{2}[(\frac{1}{2}mR^2)\omega^2] + \frac{1}{2}m(\omega R)^2\\\\E_2 = \frac{1}{2}[(\frac{1}{2}*100*0.3^2)\omega^2] +\frac{1}{2}*100(\omega)^2*0.3^2\\\\E_2 = 2.25 \omega^2 +4.5 \omega^2\\\\E_2 = 6.75 \omega^2

Recall that U = E₂

1885.2 = 6.75ω²

ω² = 1885.2/6.75

ω² = 279.2889

ω = √279.2889

ω = 16.712 rad/s

Therefore, the angular velocity of the disk when it has turned one revolution is 16.712 rad/s

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A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77
Vikki [24]
The formula for the rotational kinetic energy is

KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2}  =2.07459 \ kgm^{2}

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}

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6 0
3 years ago
Please help with this problem. Thank you.​
4vir4ik [10]

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

4 0
3 years ago
The metal gold crystallizes in a face centered cubic unit cell with one atom per lattice point. When X-rays with λ = 1.436 Å are
Umnica [9.8K]

Answer:

 r =  1.45 Å

Explanation:

given,

λ = 1.436 Å

θ = 20.62°

d = a

n = 2

metal gold crystallizes in a face centered cubic unit cell

Radius of the gold atom = ?

using Bragg's Law

 n λ = 2 d sin θ

 2 x 1.436 Å = 2 a sin 20.62°

 a = 4.077 Å

We know relation of radius for face centered cubic unit cell

 a = \dfrac{4r}{\sqrt{2}}

 4.077= \dfrac{4\times r}{\sqrt{2}}

 r =  1.45 Å

the radius of a(n) gold atom. is equal to 1.45 Å

7 0
3 years ago
A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?
sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

Q=mC\Delta T

where

m is the mass of the substance

C is the specific heat capacity

\Delta T is the change in temperature

In this problem:

m = 1 g is the mass of water

C=1 cal/g^{\circ}C is  the specific heat capacity of water

\Delta T=1^{\circ}C is the change in temperature

So, the heat needed is

Q=(1)(1)(1)=1 cal

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

Q=m\lambda_f

where

m is the mass of the substance

\lambda_f is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: m=1g

- Specific latent heat of fusion of ice: \lambda_f=80 cal/g

So, the heat needed is

Q=(1)(80)=80 cal

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

Q=m\lambda_v

where

m is the mass of the substance

\lambda_v is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: m=1g

- Specific latent heat of vaporization of water: \lambda_v=540 cal/g

So, the heat needed is

Q=(1)(540)=540 cal

5 0
3 years ago
Physics 5
Darina [25.2K]

Answer:

0.85 A

Explanation:

the effective current is 0.85 A

6 0
2 years ago
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