Just number 7a,b,c,d,e,f 7. Mary\'s merry morning ry go-round. Mary has mass m.
ID: 1782674 • Letter: J
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Just number 7a,b,c,d,e,f
7. Mary's merry morning ry go-round. Mary has mass m. She is not holding onto on is solely responsible for preventing here from falling off the merry coefficients of kinetic and static friction between Mary's shoes and the Mary is standing on a mer go-round. The surface of the merry go-round are k and , respectively. direction of the friction force when Mary is standing still relative to atic a. Describe the the surface of the merry go-round at a particular point. Is this friction force st or kinetic Mary is standing still at a distance r from the center of rotation of the merry go-round, and if the merry go-round is going 'round at an angular speed (radians b. If per second) a, what is the force of friction f on her shoes? c. Suppose that at her current position a distance r from the center of rotation, the force of friction is half its maximum. If Mary stays at that position, what's the maximum angular speed wmax the merry go-round can have before Mary's shoes start to slip? d. Suppose instead that remains constant but that Mary walks to a different dis- tance from the center. Will the force of friction increase or decrease if she walks closer to / further from the center? At what radius can she stand before her shoes start to slip? e. Suppose that Mary is standing 2 m from the center of a merry go-round. Estimate wmax. Hint. Use the value of the coefficient of static friction between your shoes and a smooth surface from the last problem. f. Will any of the answers to the parts of this problem change if mary is standing on Jarringly Jubilant Jumping mate the force you apply on the ground when you jump as high as you can one foot is problem is to lead to through a calculation that allows you to approxi- When a person of mass m jumps, she first bends her knees a certain amount so that herExplanation / Answer
a)The direction of frictional force is always apposite to the direction of other forces acting. While she is still only gravity acts on her, so its upwards since gravity acts downwards.
Its static friction.
b)The centripital force acting on Merry is:
Fc = mv^2/r
this balances the static frictional force acting on merry.
Fo, f = Fc = mv^2/r ; but v = r w
f = m r^2 w^2/R = m r w^2
Hence, f = m r w^2
c)f' = 1/2 f
For not to slip the static frictional force acting on merry should balance the centripital force.
Fc = m r w(max)^2 = f
w(max) = sqrt (f/m r)
d)We see that
f = m r w^2
the frictional force will increase as she walk away /father from the center.
e)Last Problem required
f)Yes, the contact force is frictional force and as the contact area decreases, it will also reduce and she may have more chanes of slipping ir getting thrown away.
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