{"id":182330,"date":"2025-01-13T16:42:42","date_gmt":"2025-01-13T16:42:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182330"},"modified":"2025-01-13T16:42:45","modified_gmt":"2025-01-13T16:42:45","slug":"your-cousin-throckmorton-skateboards-from-rest-down-a-curved-ramp-in-the-form-of-a-quarter-circle-of-radius-r-2-50m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/your-cousin-throckmorton-skateboards-from-rest-down-a-curved-ramp-in-the-form-of-a-quarter-circle-of-radius-r-2-50m\/","title":{"rendered":"Your cousin Throckmorton skateboards from rest down a curved ramp in the form of a quarter circle of radius R = 2.50m"},"content":{"rendered":"\n

Your cousin Throckmorton skateboards from rest down a curved ramp in the form of a quarter circle of radius R = 2.50m. Assume that the wheels of the skateboard roll without slipping and that the friction in the wheel bearings is negligible. If we treat Throcky and his skateboard as a particle of mass m = 27.0kg, he undergoes (nonuniform) circular motion as he rides down the ramp.
a) When Throcky has descended half the distance to the bottom, so that the surface of the ramp is at 45\u00b0 to the horizontal, what is his instantaneous speed?
b) What is the magnitude of the normal force acting on Throcky at that moment?<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s analyze Throckmorton’s motion down the quarter-circle ramp to determine his instantaneous speed and the magnitude of the normal force acting on him when the ramp is at a 45\u00b0 angle to the horizontal.<\/p>\n\n\n\n

Part (a): Instantaneous Speed at 45\u00b0<\/strong><\/p>\n\n\n\n

At the top of the ramp, Throcky has only gravitational potential energy, which converts into kinetic energy as he descends. At the 45\u00b0 point, half of the ramp’s height has been descended, so the change in height (\u0394h) is:<\/p>\n\n\n\n

\u0394h = R – R * cos(45\u00b0) = R * (1 – cos(45\u00b0))<\/p>\n\n\n\n

The potential energy lost equals the kinetic energy gained:<\/p>\n\n\n\n

m * g * \u0394h = 0.5 * m * v\u00b2<\/p>\n\n\n\n

Simplifying and solving for v:<\/p>\n\n\n\n

v = \u221a(2 * g * \u0394h)<\/p>\n\n\n\n

Substituting \u0394h:<\/p>\n\n\n\n

v = \u221a(2 * g * R * (1 – cos(45\u00b0)))<\/p>\n\n\n\n

Given R = 2.50 m and g \u2248 9.81 m\/s\u00b2:<\/p>\n\n\n\n

v \u2248 \u221a(2 * 9.81 * 2.50 * (1 – cos(45\u00b0)))<\/p>\n\n\n\n

v \u2248 5.00 m\/s<\/p>\n\n\n\n

Part (b): Magnitude of the Normal Force at 45\u00b0<\/strong><\/p>\n\n\n\n

At the 45\u00b0 point, the forces acting on Throcky are:<\/p>\n\n\n\n