0. Physics 1710 Chapter 6—Circular Motion. Answer: a = ∆v/∆ t a = (112 m/s)/(2.5 sec) = 44.8 m/s 2 Thrust = force F = ma = (1800. kg)(44.8 m/s 2 ) ~ 81 kN. 0. Physics 1710 Chapter 6—Circular Motion. 1 ′ Lecture:
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The net force on a body executing circular motion is equal to the mass times the centripetal acceleration of the body. Fcentripetal = m acentripetal
acentripetal = v 2/ R [toward the center]; a = -ω2r
In non-inertial frames of reference one may sense fictitious forces.
At terminal velocity the velocity-dependent resistive forces balance the accelerating forces so that no further acceleration occurs.
x = r cos θy= r sin θ
vx = (dr/dt) cos θ- r sin θ(d θ/dt)vy = (dr/dt) sin θ+r cos θ(d θ/dt)Let (dr/dt) = 0 vx = - r ω sin θvy = +r ω cos θ
ax = dvx /dt= d(- r ω sin θ)/dt = r ω2cos θ- r ω sin θ(d ω/dt)ay = dvy /dt= d(r ω cos θ)/dt = -r ω2sin θ+ r ω cosθ(d ω/dt)a = - ω2r + r (d vtangential /dt)
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An athlete swings a 8 kg “hammer” around on a 1.2 m long cable at an angular frequency of 1.0 revolution per second. How much force must he exert on the hammer throw handle?