A thin rod with mass M = 5.85 kg is bent in a semicircle of radius R = 0.606 m (the figure below). (a) What is its gravitational force (both magnitude and direction) on a particle with mass m = 3.37×10-3 kg at P, the center of curvature? (b) What would be the force on the particle if the rod were a complete circle?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

A thin rod with mass M = 5.85 kg is bent in a semicircle of radius R = 0.606 m (the figure below). (a) What is its gravitational force (both magnitude and direction) on a particle with mass m = 3.37×10-3 kg at P, the center of curvature? (b) What would be the force on the particle if the rod were a complete circle?

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
i can help later
thank you @IrishBoy123

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1442784318789:dw|
you can start by using calculus the force between the central mass M , and the small elemental section of mass dm on the rod follows Newton's Law for gravitation, ie: \(\large |dF| = \frac{GMdm}{R^2}\) the force on dm acts inwardly toward the centre of the circle we can [and need] resolve it in the y direction [in the x direction, the force is symmetrical: overall, left and right, it nets off to zero]. so \(\large dF_y = -\frac{GMdm}{R^2} \, \sin \phi\), with the minus sign because it acts down the y axis we are given the overall mass, m, of the wire so we can say that mass dm of our small element, which subtends an angle \(d\phi\), is \(\large dm = m \times \frac{ d\phi}{\pi}\), which leaves us with : \[dF_y = -\frac{GMm \times \frac{ d\phi}{\pi}}{R^2} \sin \phi = -\frac{GMm}{\pi R^2} \sin \phi \, d\phi\] thus \[F_y = -\frac{GMm}{\pi R^2} \int\limits_{0}^{\pi} \, \sin \phi \, d\phi\] \[= -\frac{GMm}{\pi R^2} \left[ -\cos \phi \right]_{0}^{\pi} = \frac{GMm}{\pi R^2} \left[ \cos \phi \right]_{0}^{\pi} \] \[= -2\frac{GMm}{\pi R^2} \] and, here, we have: m [the rod] = 5.85 kg R [radius of rod] = 0.606 m M [the central particle] = 3.37×10-3 kg remember, the - sign just tells us that the force acts down the y axis if the rod were a complete circle,clearly symmetry would rule the day. the great thing is that we can complement that thought just using the maths. in that event we would have : \[F_y = -\frac{GMm}{\pi R^2} \int\limits_{0}^{\color{red}{2\pi}} \, \sin \phi \, d\phi \color{red} {= 0}\]
|dw:1442787001551:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question