Find the energy stored in the solid flywheel of mass 8.0×104 and radius 1.8 when it's rotating at 290
Kr=1/2*(I*w^2)
I=1/2*(M*R^2)
w=(290*2pi)/60
Kr=1/2((290*2pi)/60)^2*(.5*8.0*10^4*1.8^2)
Kr=~~~(whatever that equals my question was different just be glad i did this at all ahah)
(this is all assuming you mean 10^4 and 290 is in rpm)
To find the energy stored in a solid flywheel, we can use the formula for rotational kinetic energy:
Rotational kinetic energy (KE) = (1/2) * moment of inertia * angular velocity^2
The moment of inertia for a solid cylinder is given by the formula:
Moment of inertia (I) = (1/2) * mass * radius^2
Given:
Mass (m) = 8.0 × 10^4 g (convert to kg: 8.0 × 10^4 g ÷ 1000 g/kg)
Radius (r) = 1.8 m
Angular Velocity (ω) = 290 rad/s
First, let's convert the mass to kg:
m = 8.0 × 10^4 g ÷ 1000 g/kg = 80 kg
Next, we can calculate the moment of inertia:
I = (1/2) * m * r^2
I = (1/2) * 80 kg * (1.8 m)^2
Now, calculate the rotational kinetic energy:
KE = (1/2) * I * ω^2
KE = (1/2) * [(1/2) * 80 kg * (1.8 m)^2] * (290 rad/s)^2
Simplifying the equation:
KE = (1/2) * [(1/2) * 80 * 3.24] * 84100
KE = (1/2) * 20 * 84100
KE = 10 * 84100
KE = 841,000 J
Therefore, the energy stored in the solid flywheel is 841,000 Joules.
To find the energy stored in a solid flywheel, we can use the formula for rotational kinetic energy:
E = (1/2) I ω^2
Where:
E is the rotational kinetic energy
I is the moment of inertia
ω is the angular velocity
First, let's calculate the moment of inertia. For a solid flywheel, the moment of inertia is given by:
I = (1/2) m r^2
Where:
m is the mass of the flywheel
r is the radius of the flywheel
Given:
m = 8.0 × 10^4 kg
r = 1.8 m
Plugging in the values, we can calculate I:
I = (1/2) × (8.0 × 10^4 kg) × (1.8 m)^2
= 1.296 × 10^5 kg·m^2
Next, we need to calculate the angular velocity ω in radians per second. We are given the angular velocity in revolutions per minute (rpm), so we need to convert it to radians per second. To do this, we use the following conversion:
1 revolution = 2π radians
1 minute = 60 seconds
ω = (290 rpm) × (2π rad/1 revolution) × (1/60 min/1 second)
≈ 30.33 radians/second
Now we can calculate the rotational kinetic energy E:
E = (1/2) × (1.296 × 10^5 kg·m^2) × (30.33 radians/second)^2
≈ 582.71 × 10^5 joules
Therefore, the energy stored in the solid flywheel when it's rotating at 290 rpm is approximately 582.71 × 10^5 joules.