Scientists design a new particle accelerator in which protons (mass 1.7 × 1027 k

Question

Scientists design a new particle accelerator in which protons (mass 1.7 × 1027 kg) follow a circular trajectory given by r =ccos(kt2)^+csin(kt2)^, where c = 4.5 m and k = 8.0 × 104 rad/s2 are constants and t is the elapsed time.

My question is for part C (See photo attached).

Problem 8.60 Scientists design a new particle accelerator in which protons (mass 1.7 x 10 27 kg) follow a circular trajectory given by r c cos(kt2)i c sin(kt2)j, where c 34.5 m and k 8.0 x 104 rad /s2 are constants and t is the elapsed time Part A What is the radius of the circle? Express your answer with the appropriate units 4.5 m Submit My Answers Give Up Correct Part B What is the proton's speed at t 3.0 s? Express your answer with the appropriate units 2.2x10 Submit My Answers Give Up Correct Part C What is the force on the proton at t 3.0 s? Give your answer in component form Enter your answers separated by a comma.

Explanation / Answer

r = c*cos(kt^2)i + c*sin(kt^2) j

v = dr/dt = -c*sin(kt^2)*2*kt i + c*cos(kt^2)*2*k*t j

accelerattion a = dv/dt

a = -c*2*k(t*cos(kt^2)*2kt + sin(kt^2)) i + c*2*k(-t*sin(kt^2)*2kt + cos(kt^2))j

Fx = -m*c*2*k(t*cos(kt^2)*2kt + sin(kt^2))

Fx = -1.7*10^-27*4.5*2*8*10^4*(cos(8*10^4*3^2)*2*8*10^4*3 + sin(8*10^4*3^2))

Fx = -5.47*10^-16 N

Fy = m*ay = m*c*2*k(-t*sin(kt^2)*2kt + cos(kt^2))

Fy = 1.7*10^-27*4.5*2*8*10^4*(-sin(8*10^4*3^2)*2*8*10^4*3 + cos(8*10^4*3^2))

Fy = 2.13*10^-16 N

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