A 2.10 kg block is cemented 32.0 cm from the center of a 3.40 kg circular platfo

Question

A 2.10 kg block is cemented 32.0 cm from the center of a 3.40 kg circular platform with a radius of 0.550 m. The platform is free to rotate about its center. A 0.0640 kg bullet is fired horizontally with a velocity of 320 m/s and becomes lodged in the block. (a) What is the angular velocity of the platform after the collision? (b) After the collision, you try to stop the platform from spinning by pushing on the side of the platform with a force of 4.25 N. Assuming there is a coefficient of friction between your finger and the platform of 0.223, how long does it take the platform to stop moving?

Iblock=mr^2

Iplatform=(1/2)mr^2

I know the answers are:

8a. 5.67 rad/s
8b. 12.6 s

I don't know how they got them.

Explanation / Answer

8.9 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.