{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW%206 - MasteringPhysics HW#6 Ch 9 8.5 Model We will use...

This preview shows pages 1–3. Sign up to view the full content.

MasteringPhysics HW #6 Ch 9 8.5. Model: We will use the particle model for the car which is in uniform circular motion. Visualize: Solve: The centripetal acceleration of the car is a r = v 2 r = 15 m/s ( ) 2 50 m = 4.5 m/s 2 The acceleration is due to the force of static friction. The force of friction is f s = ma r = 1500 kg ( ) 4.5m s 2 ( ) = 6750 N = 6.8 kN. Assess: The model of static friction is f s ( ) max = n μ s = mg μ s mg 15,000 N since μ s 1 for a dry road surface. We see that f s < f s ( ) max , which is reasonable. 8.11. Model: The satellite is considered to be a particle in uniform circular motion around the moon. Visualize: Solve: The radius of the moon is 1.738 × 10 6 m and the satellite’s distance from the center of the moon is the same quantity. The angular velocity of the satellite is ω = 2 π T = 2 π rad 110 min × 1min 60 s = 9.52 × 10 4 rad/s and the centripetal acceleration is a r = r ω 2 = 1.738 × 10 6 m ( ) 9.52 × 10 4 rad/s ( ) 2 = 1.58 m/s 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The acceleration of a body in orbit is the local “ g ” experienced by that body. 8.23. Model: Treat the motorcycle and rider as a particle. Visualize: This is a two-part problem. Use an s -axis parallel to the slope for the first part, regular xy - coordinates for the second. The motorcycle’s final velocity at the top of the ramp is its initial velocity as it becomes airborne. Solve: The motorcycle’s acceleration on the ramp is given by Newton’s second law: ( F net ) s = f r mg sin20 ° = μ r n mg sin20 ° = μ r mg cos20 ° mg sin20 ° = ma 0 a 0 = g ( μ r cos20 ° + sin20 ° ) = (9.8 m/s 2 )((0.02)cos20 ° + sin20 ° ) = 3.536 m/s 2 The length of the ramp is s 1 = (2.0 m)/sin20 ° = 5.85 m. We can use kinematics to find its speed at the top of the ramp: v 1 2 = v 0 2 + 2 a 0 ( s 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern