n an amusement park Dave (m=80kg) is enjoying himself in a ROTOR.The rotor has a

Question

n an amusement park Dave (m=80kg) is enjoying himself in a ROTOR.The rotor has a radius of 2m, and rotates around its axis with aperiod of 1.5s. he is pressed against the wall and the floor hasdisappeared. the coefficients of friction between dave and the wallare MK=.3, Ms = .5
a. what is the magnitude and direction of the acceleration ofdave.
b. what is the centripetal force that provides the acceleration?what is its magnitude?
c. what is the magnitude and direction of the force offriction?
d. now the rotor reduce its speed. how slow can the speed at therim minimally be, before Dave begins to slide downwards.?

Explanation / Answer

w = 2/T = 4.19 rad /sec a) there is only radial acceleration = F/m = m*w^2*r /m =w^2*r =35 rad /sec^2 outwards in the radial direction b) F= m*w^2*r =2809 N c) the centripital force here is like normal force forfriction Friction = N* = 2809*0.5 =1404.5 N upwards d) so friction is = m*w^2*r* and opposite force ismg m*w^2*r* = m*g where = 0.3 solve for w= 4.04 rad /sec this is minimum

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.