Question 1
A bowling ball has a mass of 4.5 kg, a moment of inertia of 0.072 kg · m2, and a radius of 0.2 m. It rolls along the lane without slipping at a linear speed of 4.1 m/s.
What is the total kinetic energy of the rolling ball?
Answer in units of J
i got some answer betwen 50 and keep getting it wrong
Question 2
A coin with a diameter of 2.7 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 25.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of magnitude 2.47 rad/s2,
How far does the coin roll before coming to rest?
Final question
A solid, horizontal cylinder of mass 5.6 kg and radius 0.92 m rotates with an angular speed of 2.2 rad/s about a fixed vertical axis through its center. A 0.245 kg piece of putty is dropped vertically onto the cylinder at a point 0.585 m from the center of rotation, and sticks to the cylinder.
What is the final angular speed of the sys- tem?
Answer in units of rad/s
A bowling ball has a mass of 4.5 kg, a moment of inertia of 0.072 kg · m2, and a radius of 0.2 m. It rolls along the lane without slipping at a linear speed of 4.1 m/s.
What is the total kinetic energy of the rolling ball?
Answer in units of J
i got some answer betwen 50 and keep getting it wrong
Question 2
A coin with a diameter of 2.7 cm is dropped onto a horizontal surface. The coin starts out with an initial angular speed of 25.4 rad/s and rolls in a straight line without slipping. If the rotation slows with an angular deceleration of magnitude 2.47 rad/s2,
How far does the coin roll before coming to rest?
Final question
A solid, horizontal cylinder of mass 5.6 kg and radius 0.92 m rotates with an angular speed of 2.2 rad/s about a fixed vertical axis through its center. A 0.245 kg piece of putty is dropped vertically onto the cylinder at a point 0.585 m from the center of rotation, and sticks to the cylinder.
What is the final angular speed of the sys- tem?
Answer in units of rad/s
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K = 1/2 m v^2 + 1/2 I omega^2 = 1/2( 4.5) 4.1^2 + 1/2 (0.072) (v/r = 4.1/.2)^2
= I got 52.9 J
omegaf^2 = omegai^2 + 2 alpha deltatheta
0 = 25.4^2 + 2 (-2.47) deltatheta so deltatheta = 130.6 rads
130.6 rads = 130.6/2Pi revs = 20.8 revs
One rev = circumference = 2 Pi r = 2 Pi 0.0135 m = 0.085 m
distance = revs x circumference = 1.76 m
I1 omega1 = I2 omega2
1/2 M r^2 omega1 = (1/2 M r^2 + m2 r2^2 ) omega2
1/2 (5.6)(.92^2) = 2.37
2.37 x 2.2 = (2.37 + 0.245 x 0.585^2) omega2 solve for omega2
I got 2.12 rad/s (sounds about right)
= I got 52.9 J
omegaf^2 = omegai^2 + 2 alpha deltatheta
0 = 25.4^2 + 2 (-2.47) deltatheta so deltatheta = 130.6 rads
130.6 rads = 130.6/2Pi revs = 20.8 revs
One rev = circumference = 2 Pi r = 2 Pi 0.0135 m = 0.085 m
distance = revs x circumference = 1.76 m
I1 omega1 = I2 omega2
1/2 M r^2 omega1 = (1/2 M r^2 + m2 r2^2 ) omega2
1/2 (5.6)(.92^2) = 2.37
2.37 x 2.2 = (2.37 + 0.245 x 0.585^2) omega2 solve for omega2
I got 2.12 rad/s (sounds about right)