A 587 kg roller coaster car (includes mass of occupants) are passing through a vertical loop. The speed of the car at the top of the loop is 15.4 m/s. What radius of curvature (in meters) must the loop have at its very top in order for the occupants to experience a normal force which is 1/3 their weight?

Two students sitting in adjacent seats in a lecture room have weights of 602 N and 682 N. Assume that Newton's law of universal gravitation can be applied to these two students and find the gravitational force (in Newtons) that one student exerts on the other when they are separated by .596 m. Express your answer using scientific notation.

How many Earth radii above the Earth (not from its center) must you be located to experience an acceleration of gravity of 1.952 m/s/s. Express in terms of Earth-radii; that is, express the answer as the number of times greater than 6.38 x 10^6 m. Enter your answer to the third decimal place.

Solution Summary

This solution has step-by-step calculations and also annotated diagrams to answer the circular motion and weight questions. All formulas used are shown clearly.

... The net downwards force is now reduced (weight upwards resistive force ... In case of circular motion, the change in momentum is directed towards the centre of ...

... acceleration which is changing the direction of motion just enough to keep it in a circular path with ... 2. Why is it easier to hold a 10lb weight next to ...

... (Note: Ignore the weight of the board) A. What ... The solution set includes various problems on circular motion, such as, angular speed, acceleration, and force. ...

... e. It is the point from which the torque produced by the weight of the ... This set of multiple choice problems on circular and rotational motion, centre of ...

... The vertical component ie N cos balances the weight of the vehicle ... the center of the curve provides the necessary centripetal force for the circular motion. ...

... L θ Vv T During the subsequent motion at some time the rope ... acting on the man θ mg 1. his weight mg acting ... in the rope towards the center of the circular path ...

... Express your answer in terms of the pilot's weight mg. 72. A student investigating circular motion places a dime 10 cm from the center of a 33 1/3 -rpm record ...

... satellites are placed in a circular orbit that ... to answer each question regarding Newton's Law of Motions. 1. ... appropriate one of Newton's three laws of motion. ...

... ie Potential energy at C) (2) EKC - To continue the circular motion (ie Kinetic ...Weight of the skier = Centripetal force ie ie Substituting in equation (2), we ...

... So the reading ( N ) is your true weight (W = mg ... 2 11 Option D is correct because in a uniform circular motion, the force acting on the body is cetripetal force ...