A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?

Answers

Answer 1

Answer:

To solve this problem it is necessary to apply the related concepts to the moment of inertia in a disk, the conservation of angular momentum and the kinematic energy equations for rotational movement.

PART A) By definition we know that the moment of inertia of a disk is given by the equation

$I = (1)/(2) MR^2$

Where

M = Mass of the disk

R = Radius

Replacing with our values we have

$I = (1)/(2) (0.2)(0.3)^2$

$I = 9*10^(-3)kg\cdot m^2$

The initial angular momentum then will be given as

$I = I_1 \omega_1$

$I = 0.012*2$

$I = 0.024kg\cdot m^2/s$

Therefore the total moment of inertia of the table and the disc will be

$I_2 = 9*10^(-3)+0.012$

$I_2 = 0.021kg\cdot m^2$

The angular velocity at the end point will be given through the conservation of the angular momentum for which it is understood that the proportion of inertia and angular velocity must be preserved. So

$I_1 \omega_1 = I_2\omega_2$

$(0.012)(2)=(1.08*10^(-4))\omega_2$

$\omega_2 = (0.012*2)/(0.021)$

$\omega_2 = 1.15rad/s$

Therefore the new angular velocity is 1.15rad/s

PART B) Through the conservation of rotational kinetic energy we can identify that its total change is subject to

$\Delta KE = (1)/(2)I_1\omega_1^2-(1)/(2)I_2\omega^2$

$\Delta KE = (1)/(2)(I_1\omega_1^2-I_2\omega^2)$

$\Delta KE = (1)/(2)(0.024*2^2-0.021*1.15^2)$

$\Delta KE = 0.034J$

Therefore the change in kinetic energy is 0.034J

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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 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

$KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)$

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,

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

$KE_(rot) =18,671.31 \ J$

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> $573*2\pi *(1)/(60) =60$ rad/s

I= $(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537$ kgm^2

K=1/2(number of rods)(I)(ω)= $(1)/(2) *(5)(0.537)(60)^2=4833$ J

I know it's very late, but hope this helps anyone else trying to find the answer.

What is required for a sound wave to be reflected?

Answers

when sound travels on a certain wave pattern or medium once it hits the surface or surfaces of another wave or mdeium and then it bounces back that is a reflected sound wave.

A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse

Answers

The amount of work done per second by the horse exerting a force of 1800 N on a wagon moving with a speed of 0.4 m/s is 720 J/s.

What is power?

Power is the workdone by a body in one second.

To calculate the work done by the horse in one seconds, we use the formula below

Formula:

P = Fv................ Equation 1

Where:

P = work done on the horse in one second
F = Force of the horse
v = Velocity of the wagon

From the question,

Given:

F = 1800 N
v = 0.4 m/s

Substitute these values into equation 1

P = 1800×0.4
P = 720 J/s

Hence, the amount of work done per second by the horse is 720 J/s.

Learn more about power here: brainly.com/question/25864308

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Complete question: A wagon is pulled at a speed of 0.40 m/s by a horse exerting 1800 Newtons of horizontal Force. how much work was done by the horse per second.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

Answers

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

James weighs 120 pounds. He ran up the steps which rose 12 feet. John weighs 91 pounds. He ran up the same steps as James. If both boys reached the top of the 12-foot stairs at the same time, which boy had the greatest horsepower?

Answers

Mass of James = 54.431kg
Distance ran by James = 3.6576m
Mass of John= 41.277kg
Distance relan by John= 3.6576m

Work done by James = 54.431kg*10N/kg*3.6576m= 1990.87J

Work done by John= 41.277kg*10N/kg*3.6576m=1509.75J

Horsepower = work done / time since time is same, James has larger horsepower

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