A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.72 m. The puck makes one revolution every t = 0.45 s. Part (a) What the magnitude of the tension in the string, in newtons, while the puck revolves?

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in radial direccion (N)

m : puck mass (kg)

a : radial acceleration of the puck (m/s²)

dа ta:

m = 0.085 kg

L = 0.72 m = R : radium of the circular path (m)

θ = one revolution = 2Π rad

t = 0.45 s

Angular speed of the puck

ω = θ/t

ω = 1 rev/0.45 s = (2π/0.45) rad/s

ω = 13.96 rad / s

Radial acceleration or centripetal

a = ω²*R

a = (13.96) ² * (0.72)

a = 140.3 m/s²

Magnitude of the tension in the string (T)

We apply the Formula (1)

∑F = m*a

T = (0.085 kg) * (140.3 m/s²)

T = 11.93 N