Answer to Question #197178 in Physics for gibson

Question #197178

1.A circular race track of radius 500m is banked at angle of 15o. Calculate the maximum

permissible speed to avoid slipping if

(a) there is no friction between the tires and the road.

(b) the coefficient of friction between the tires and the road is 0.3.

2. A stone is attached to the end of a rope of length l = 25 cm, and rotates the stone at a

constant speed in a circular path. If the period of rotation is 5.0s, find the stone’s centripetal

acceleration.


1
Expert's answer
2021-05-25T10:09:51-0400

1.

(a)


"N\\cos15\u00b0=mg"

"N\\sin15\u00b0=mv^2\/R"


So, we have


"N=mg\/\\cos15\u00b0"


"mg\\sin15\u00b0\/\\cos15\u00b0=mv^2\/R\\to v=\\sqrt{gR\\tan15\u00b0}=\\sqrt{9.8\\cdot500\\cdot\\tan15\u00b0}=36.2\\ (m\/s)"


(b)


"N\\cos15\u00b0-\\mu N\\sin15\u00b0=mg\\to N=\\frac{mg}{\\cos15\u00b0-\\mu \\sin15\u00b0}"

"N\\sin15\u00b0+\\mu N\\cos15\u00b0=mv^2\/R"


"\\frac{mg}{\\cos15\u00b0-\\mu \\sin15\u00b0}(\\sin15\u00b0+\\mu \\cos15\u00b0)=mv^2\/R"


"v=\\sqrt{\\frac{gR}{\\cos15\u00b0-\\mu \\sin15\u00b0}(\\sin15\u00b0+\\mu \\cos15\u00b0)}="


"=\\sqrt{\\frac{9.8\\cdot500}{\\cos15\u00b0-0.3\\cdot \\sin15\u00b0}(\\sin15\u00b0+0.3\\cdot \\cos15\u00b0)}=55\\ (m\/s)"


2.


"a_n=\\omega^2r=(\\frac{2\\pi}{T})^2r=(\\frac{2\\cdot3.14}{5})^2\\cdot0.25=0.39\\ (m\/s^2)"








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