An object of mass 1.00kg is moving over a horizontal circular path of radius 2.00m, with a speed of 3.00ms^-1. Determine:
(a) its centripetal acceleration
(b) its kinetic energy
(c) its period
(d) its angular velocity
To determine the answers to these questions, we will use the following formulas:
(a) Centripetal acceleration:
Ac = v^2 / r
where Ac is the centripetal acceleration, v is the velocity, and r is the radius of the circular path.
(b) Kinetic energy:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
(c) Period:
T = 2π * r / v
where T is the period, r is the radius, and v is the velocity.
(d) Angular velocity:
ω = v / r
where ω is the angular velocity, v is the velocity, and r is the radius.
Let's plug in the given values to solve for each question.
(a) Centripetal acceleration:
Ac = (3.00ms^-1)^2 / 2.00m
Ac = 9.00m^2s^-2 / 2.00m
Ac = 4.50m/s^2
Therefore, the centripetal acceleration is 4.50m/s^2.
(b) Kinetic energy:
KE = (1/2) * 1.00kg * (3.00ms^-1)^2
KE = 0.5 * 1.00kg * 9.00m^2s^-2
KE = 4.50J
Therefore, the kinetic energy is 4.50J.
(c) Period:
T = 2π * 2.00m / 3.00ms^-1
T = 2π * 2.00m / 3.00s^-1
T = 12.57s
Therefore, the period is 12.57s.
(d) Angular velocity:
ω = 3.00ms^-1 / 2.00m
ω = 1.50s^-1
Therefore, the angular velocity is 1.50s^-1.