1)Suppose a 2.4kg book were dropped and in free fall. What would be the apparent

Question

1)Suppose a 2.4kg book were dropped and in free fall. What would be the apparent weight (defined as the force required to support the object) of the book? Assume the classroom is on Earth.

2)What would be the apparent weight of the book if it were in orbit on the International Space Station?

3)What would it' apparent weight be if it were moving at 2.5m/s on a 5.0m-diameter Ferris Wheel at the low point in its motion?

4)What would its apparent weight be if it were moving at 2.5m/s on a 5.0-m diameter Ferris Wheel at the high point of its motion?

5)Suppose the textbook (2.4kg) took a ride on the Rotor, that amusement park ride in which objects stand next to the wall of a large cylinder and it begins to spin. The book is supported on the floor of the Rotor and its leaning against the wall. Suppose the rotor has a radius of 2.5m. And it spins at 33 RPM (revolutions per minute). The relevant platform we are considering is the Rotor wall, not the Rotor floor. And the apaprent weight of the book will b the normal force exerted by the Rotor on the book to hold it in place while it is spinning. Determine that normal force, and thus determine the apparent weight of the textbook.
5A)Apparent weight when pinned to the wall of a spinning Rotor?

5B) How many g-forces does this apparent weight correspond to?

6) Suppose the coefficient of static friction for the book against the Rotor wall is 0.40. Is the force of friction sufficient to hold it against the wall without a support from below, i.e. can the floor of the Rotor be lowered without causing the book to also drop? (Show work)

Explanation / Answer

1) zero as no force is stopping it from free fall.

2) again zero as in the orbit also it is freely falling

3) Normal force = N = apparent weight

N-weight = centripetal acceleration

N = mv^2/r + mg = 2.4(2.5^2 / 5 + 9.8) = 26.52N

4) mg-N=mv^2/r

N=mg - mv^2/r = 2.4(9.8-2.5^2/5) = 20.52N

5) 33rpm = 33/60 * 2pi rad/s = 3.456 rad/s

v= wr = 3.456*2.5=8.64m/s

N wall = mv^2/r=71.66 N

N floor = mg=23.52N

net normal force = apparent weight = sqrt(N wall^2 + N floor^2)

= sqrt(71.66^2+23.52^2)= 75.42N

5A) when pinned to wall the whole force is exerted by the wall

but it is still 75.42 N

5B) 75.42N=75.42/9.8 gforce = 7.7 gforce

6) friction force that can be supplied by wall = 0.40 * N wall = 0.4*71.66 = 28.66N

which is greated than the weight 23.52N

so it can be supported by the friction

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