Question

A bowling ball is rolling (without slipping) across a horizontal surface. the mass of the ball is 6 kg and the radius is 10.8 cm. the initial speed of the ball is 8 m/s. use this information to answer the next 4 questions.

A) what is the moment of inertia of the bowling ball?
B) what is the inertia translational kinetic energy of the bowling ball?
C) what is the initial rotational kinetic energy of the bowling ball?
D) the bowling ball comes to the bottom of a ramp that is inclined 20 degrees with respect to the horizontal. what maximum height will the ball reach of this ramp?

Answer 1

A) What is the moment of inertia of the bowling ball?

The moment of inertia of a bowling ball is 2/5mr

2

, where m is the mass of the ball and r is the radius. In this case, m=6 kg and r=10.8 cm=0.108 m. Therefore, the moment of inertia of the bowling ball is:

I=

5

2

​

(6 kg)(0.108 m)

2

=0.023328 kg m

2

B) What is the translational kinetic energy of the bowling ball?

The translational kinetic energy of a bowling ball is

2

1

​

mv

2

, where m is the mass of the ball and v is the velocity. In this case, m=6 kg and v=8 m/s. Therefore, the translational kinetic energy of the bowling ball is:

KE

t

​

=

2

1

​

(6 kg)(8 m/s)

2

=192 J

C) What is the initial rotational kinetic energy of the bowling ball?

The initial rotational kinetic energy of a bowling ball is

2

1

​

Iω

2

, where I is the moment of inertia of the ball and ω is the angular velocity. In this case, I=0.023328 kg m

2

and ω=v/r=8 m/s/0.108 m=74.074 rad/s. Therefore, the initial rotational kinetic energy of the bowling ball is:

KE

r

​

=

2

1

​

(0.023328 kg m

2

)(74.074 rad/s)

2

=119.36 J

D) The bowling ball comes to the bottom of a ramp that is inclined 20 degrees with respect to the horizontal. What maximum height will the ball reach up this ramp?

The maximum height that the ball will reach up the ramp can be found using the following equation:

h=

2g

v

2

​

sin

2

(θ)

where v is the initial velocity of the ball, g is the acceleration due to gravity, and θ is the angle of the ramp. In this case, v=8 m/s, g=9.8 m/s

2

, and θ=20

∘

. Therefore, the maximum height that the ball will reach up the ramp is:

h=

2(9.8 m/s

2

)

(8 m/s)

2

​

sin

2

(20

∘

)=1.53 m

Step-by-step explanation:

Answer 2

A) The moment of inertia of the bowling ball is 0.139 kg*m^2. B) The translational kinetic energy of the bowling ball is 192 J. C) The initial rotational kinetic energy of the bowling ball is 295.24 J. D) The ball will reach a maximum height of 6.18 meters on the ramp.

Step-by-step explanation:

To answer these questions, we need to use the formula for the moment of inertia of a solid sphere, which is given by I = (2/5)*m*r^2, where m is the mass of the sphere and r is the radius.

A) The moment of inertia of the bowling ball can be calculated as I = (2/5)*m*r^2 = (2/5)*(6 kg)*(0.108 m)^2 = 0.139 kg*m^2.

B) The translational kinetic energy of the bowling ball can be calculated as KE = (1/2)*m*v^2, where v is the velocity of the ball. Plugging in the values, we get KE = (1/2)*(6 kg)*(8 m/s)^2 = 192 J.

C) The initial rotational kinetic energy of the bowling ball can be calculated as KER = (1/2)*I*ω^2, where ω is the angular velocity of the ball. Since the ball is rolling without slipping, ω = v/r. Plugging in the values, we get KER = (1/2)*(0.139 kg*m^2)*((8 m/s)/(0.108 m))^2 = 295.24 J.

D) To find the maximum height the ball will reach on the ramp, we need to use the conservation of mechanical energy. The initial mechanical energy of the ball is the sum of its initial translational kinetic energy and initial rotational kinetic energy. And the final mechanical energy of the ball at the maximum height is the sum of its potential energy and final rotational kinetic energy (which would be 0 since the ball is at rest). So we can set the initial mechanical energy equal to the final mechanical energy and solve for the maximum height. Plugging in the values, we get (1/2)*(6 kg)*(8 m/s)^2 + (1/2)*(0.139 kg*m^2)*((8 m/s)/(0.108 m))^2 = (6 kg)*9.8 m/s^2*h, where h is the height. Solving for h, we find that the ball will reach a maximum height of 6.18 meters on the ramp.