Question

Yesterday, I went bowling and while I was waiting for my turn, I decided to compute the total kinetic energy of a rolling ball. Treat the ball as a solid sphere with a diameter of 22.0 cm and a mass of 5.00 kg, the ball rolls down the lane without slipping at a speed of 15.0 km/h. What did I find for the total kinetic energy

Answer 1

Ktot = 60.8 J

Step-by-step explanation:

• The total kinetic energy of a rolling ball, is composed by a rotational kinetic energy (due to the rotation around a diameter), plus a traslational kinetic energy, due to the translation of the center of mass.
• The rotational kinetic energy can be written as follows:

`K_(rot)= (1)/(2)* I * \omega^(2) (1)`

• where I = moment of Inertia of a solid sphere, and ω², the square of the angular speed.
• If the ball rolls without slipping, there exists a fixed relationship between the angular speed and the speed of the center of mass, as follows:

`\omega = (v_(CM) )/(R) (2)`

• The traslational part, can be expressed as follows:

`K_(trasl) = (1)/(2) * m * v_(CM) ^(2) (3)`

• For a solid sphere, like a bowling ball, the moment of inertia about an axis passing through a diameter can be written as follows:

`I =(2)/(5) * m * R^(2) (4)`

• Replacing (2) and (4) in (1), we have:

`K_(rot)= (1)/(2)* (2)/(5) * m * R^(2) * ((v_(CM) )/(R)) ^(2) = (1)/(5) * m* v_(CM) ^(2) (5)`

• The total kinetic energy is just the sum of (3) and (5):

`K_(tot) = (1)/(5) * m* v_(CM) ^(2) + (1)/(2) * m * v_(CM) ^(2) = (7)/(10) * m* v_(CM) ^(2) (6)`

• The speed of the center of mass is given in km/h, so it is needed to convert it to m/s, as follows:

`v_(CM) = 15.0 (km)/(h) * (1000m)/(1km) * (1h)/(3600s) = 4.17m/s (7)`

• Replacing (7) and m=5.00 kg in (6), we get the total kinetic energy of the rolling ball, as follows:

`K_(tot) = (7)/(10) * 5.00 kg * (4.17 m/s) ^(2) = 60.8 J`