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A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24 g, with g=9.80" m"//"s"^(2). A 0.567 g coin rests on the customer's knee. Once the motion begins and in unit-vector notation, (a) What is the coin's acceleration relative to the ground? |
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Answer» Solution :The customer sitting and the coin on his knee both are in the noninertial frame of reference as the compartment is accelerating downward. Thus the concept of pseudo force comes into picture and we use this concept here. Calculation: The coin undergoes free fall. THEREFORE, with respect to GROUND, its acceleration ISS `vec(a)_("coin")=vec(g)=(-9.8" m"//"s"^(2))hatj`. (b) What is the coin.s acceleration relative to the customer? Calculation: Since the customer is being pulled down with an acceleration of `vec(a)_("customer")=1.24vec(g)=(-12.15" m"//"s"^(2))hatj`, the acceleration of the coin with respect to the customer is `vec(a)_("rel")=vec(a)_("customer")` `=(-9.8" m"//"s"^(2))hatj-(-12.15"m"//"s"^(2))hatj` `=(+2.35" m"//"s"^(2))hatj`. (c) How long does the coin take to reach the compartment ceiling, 2.20 m above the knee? Calculation: The time it takes for the coin to reach the ceiling is `t=sqrt((2h)/(a_("rel")))=sqrt((2(2.20" m"))/(2.35" m"//"s"^(2)))=1.37" s."` (d) In unit-vector notation, what is the actual force on the coin? Calculation: Since gravity is the only force acting on the coin, the actual force on the coin is `vec(F)_("coin")=m vec(a)_("coin")=m vec(g)` `=(0.567xx10^(-3)" kg")(-9.8" m"//"s"^(2))hatj` `=(-5.56xx10^(-3)" N")hatj`. (e) In unit-vector notation, what is the apparent force according to the customer.s measure of the coin.s acceleration? Calculation: In the customer.s frame, the coin travelsupward at a CONSTANT acceleration. Therefore, the apparent force on the coin is `vec(F)_("app")=m vec(a)_("rel")` `=(0.567xx10^(-3)" kg")(+2.35" m"//"s"^(2))hatj` `=(+1.33xx10^(-3)" N")hatj`. |
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