{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution #3 fall 2007

# Solution #3 fall 2007 - Physics 2211 Fall 2007 Quiz#3...

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

Physics 2211 Quiz #3 Solutions Fall 2007 g Earth = 9 . 8 m / s 2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I . (16 points) A ball of mass m is placed at a height L in the frictionless hollow cone shown. It is put into uniform circular motion. What angular velocity, ω , is required to keep the ball at the height L ? Express your answer in terms of any or all of m , L , θ , and physical or mathematical constants. ( On Earth. ) . . . . . . . . . . . . . . . . . . . . . . . Use Newton’s Second Law. The vertical, or z , direction, will yield an expression for the normal force. X F z = n z - w = ma z = 0 n sin θ/ 2 = mg n = mg sin( θ/ 2) In the radial, or r , direction, the acceleration is a centripetal acceleration. X F r = n r = ma r = m v 2 R n cos( θ/ 2) = mRω 2 Note that R/L = tan( θ/ 2), so R = L tan( θ/ 2). Substituting this expression for R and the expression for n above n cos( θ/ 2) = mRω 2 mg sin( θ/ 2) cos( θ/ 2) = m [ L tan( θ/ 2)] ω 2 mg tan( θ/ 2) = m [ L tan( θ/ 2)] ω 2 Solving for ω , ω = p g/L tan( θ/ 2) Quiz #3 Solutions Page 1 of 6

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

View Full Document
II . (16 points) A ferry boat travels from point A to a dock B across the river in 40 s. The width of the river is 200 m and the dock B is down stream. The river flows at 2 m/s. The speed of the ferry through the water is 10 m/s. What is the velocity of the ferry ( i.e. , magnitude and direction, θ ) relative to the shore? . . . . . . . . . . . . . . . . . . . . . . . The velocity of the ferry through the water, ~v fw , which has a magnitude of 10 m/s, can be resolved into two components, one across the river, v fw A , and one down the river, v fw D . The value of the component across the river is v fw A = D/ Δ t = 200 m / 40 s = 5 m / s The value of the component down the river can be found with the Pythagorean Theorem. v 2 fw = v 2 fw A + v 2 fw D v fw D = q v 2 fw - v 2 fw A v fw D = q (10 m / s) 2 - (5 m / s) 2 = 8 . 7 m / s The velocity of the ferry with respect to the shore, ~v fs is the sum of the velocity of the ferry with respect to the water, ~v fw , and the velocity of the water with respect to the shore, ~v ws .
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