sapirngieamng ltural College PHYS 1111 -Fall17-SEAT Activities and Due Dates Rot

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sapirngieamng ltural College PHYS 1111 -Fall17-SEAT Activities and Due Dates Rotational kinematics and ce /21/2017 11:00 PM 4 55.8/100 11/14/2017 01:37 PM Print CalculatorPeriodic Table ion 7 of 29 Sapling Learning The Gravitron is an amusement park ride in which riders stand against the inner wall of a large spinning steel cylinder At some point, the floor of the Graviton drops out, instilling the fear in riders that they will fall a great height. However the spinning motion of the Gravitron allows them to remain safely inside the ride. Most Gravitrons feature vertical walls, but the example shown in the figure has tapered walls of 25.4° According to knowledgeable sources, the coefficient of static friction between typical human clothing and steel ranges between 0.240 to 0.390. In the figure, the center of mass of a 54.6 kg rider resides 3.00 m from the axis of rotation. As a safety expert inspecting the safety of rides at a county fair, you want to reduce the chances of injury. What minimum rotational occupants from sliding down the wall during the ride? (expressed in revis) is needed to keep the Number - 632 revis What is the maximum rotational speed at which the riders will not slide up the walls of the ride? Number .87 rev/s O Previous Gvo up & View Saation Check Answer O Neri a E

Explanation / Answer

For maximum rotational speed (2ndpart)

Downslope gravitational acceleration:

a_g = gcos = 9.8m/s² * cos25.4º = 8.85 m/s²

Normal acceleration:

a_n = gsin + ²rsin = 9.8m/s² * sin25.4º + ²*3m*cos25.4º, so

a_n = 4.2m/s² + ² * 2.71m

Then the friction acceleration (use the lower coefficient for safety) is

a_f = 0.240 * a_n = 1.008m/s² + 0.65m * ²

and this, too, points downslope.

Upslope, we have a component of the centripetal acceleration:

a_c = ²rsin = ² * 3m * sin25.4º = ² * 1.2868m

Balancing these accelerations, we have

² * 1.2868m = 1.008m/s² + 0.65m * ² + 8.85 m/s²

Gather like terms:

² * (1.2868 - 0.65)m = (1.008 + 8.85)m/s²

² = 9.858m/s² / 0.6368m = 15.48 rad/s²

= 3.93 rad/s * 1rev/2rads = 0.626 rev/s

For the minimum, friction points upslope!

Balancing the accelerations, we have

(² * 1.2868m) + (1.008m/s² + 0.65m * ²) = 8.85 m/s²

Gather like terms:

² * (1.2868 + 0.65)m = (8.85 – 1.008)m/s²

² = 7.842m/s² / 1.9368m = 4.05 rad/s²

= 2.01 rad/s * 1rev/2rads = 0.32 rev/s

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