A car has a flywheel ( a diks of R=0.2m and uniformly distributed mass M=100kg) that can convert 25% of the rotational kinetic energy into translational kinetic energy. The mass of the car is 1000kg ( including flywheel). Suppose the car is at rest, and the flywheel has a angular speed of 200 rad/s. After all the rotational energy is converted to kinetic energy of the car, what is the speed of the car? Ignore air resistance.
A. 0 m/s; B. 3 m/s; C. 4.5 m/s; D. 9 m/s. Thanks for your help.
KE(rot)=Iω²/2 =MR²ω²/2•2
KE(tr)=0.25•KE(rot)
KE(tr)=0.25•MR²ω²/4
mv²/2 = MR²ω²/16
v=(Rω/4)•sqrt{2M/m)=4.5 m/s
C. 4.5 m/s
To solve this problem, we need to calculate the initial rotational kinetic energy of the flywheel and then determine the final translational kinetic energy of the car.
Let's start by calculating the initial rotational kinetic energy of the flywheel, which can be defined as:
K_rot = (1/2)Iω^2,
where K_rot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular speed.
The moment of inertia for a disk is given by the formula:
I = (1/2)MR^2,
where M is the mass of the flywheel and R is its radius.
Plugging in the values:
I = (1/2)(100 kg)(0.2 m)^2 = 2 kg·m^2.
Now we can calculate the initial rotational kinetic energy:
K_rot = (1/2)(2 kg·m^2)(200 rad/s)^2 = 40,000 J.
Next, we need to determine the final translational kinetic energy of the car. We know that 25% of the rotational kinetic energy is converted into translational kinetic energy. Therefore, only 25% of 40,000 J will contribute to the car's kinetic energy.
K_trans = (25/100) * K_rot
= (25/100) * 40,000 J
= 10,000 J.
Now, we can calculate the final speed of the car using the equation:
K_trans = (1/2)mv^2,
where K_trans is the translational kinetic energy, m is the mass of the car, and v is the final speed of the car.
Plugging in the values:
10,000 J = (1/2)(1000 kg)(v^2).
Simplifying the equation:
v^2 = (10,000 J * 2) / (1000 kg)
v^2 = 20 m^2/s^2.
Taking the square root of both sides:
v = √20 m/s ≈ 4.5 m/s.
Therefore, the final speed of the car after the conversion of rotational kinetic energy is approximately 4.5 m/s.
Hence, the answer is C. 4.5 m/s.