A thin, 100 g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the speed of a point on the rim

Given Information:

kinetic energy = E = 0.15 J  

mass = m = 100g = 0.1 kg

diameter = d = 8.0 cm = 0.08 m

Required Information:

Speed = v = ?

v = 2.44 m/s

Explanation:

First convert diameter into radius

r = d/2 = 0.08/2 = 0.04 m

The moment of inertia of disc is given by

I = 0.5mr²

Where m is the mass and r is the radius of the disk

I = 0.5(0.1)(0.04)²

I = 0.00008 kg.m²

and since we are dealing with rotational motion then rotational kinetic energy is given by

E  =  0.5 I ω²

we have to separate ω

ω² = E/0.5 I

ω = √E/0.5 I

ω = √0.15/0.5(0.00008)

ω = 61.23 rad/sec

Finally we know that speed is given as

v = ω r

v = 61.23 (0.04)

v = 2.44 m/s

Therefore, the speed of a point on the rim  is 2.44 m/s

The speed of a point on the rim is equal to 2.45 meters per seconds.

Given the following data:

Conversion:

Diameter of disk = 8 cm to m = = 0.08 meters

Mass of disk = 100 g to kg = = 0.1 kg

Radius = = 0.04 m

To find the speed of a point on the rim:

First of all, we would determine the moment of inertia (I) of the disk.

Moment of inertia, I = 0.00008

Next, we would solve for angular velocity from the rotational kinetic energy of the disk:

Angular velocity, w = 61.24 rad/s

Now, we can find the speed:

Speed = 2.45 m/s

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the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

gravity

explanation: