A hammer is accelerated at 13 rad/s^2 at a radiius of rotation of 1.5 m. the tangential hammer speed is 20 m/s.

a)Determine the angular velocity of the hammer
b)the magnitude of the radial component of its linear acceleration

c)the magnitude of the tangential component of its linear acceleration

a. C = pi*2r = 3.14 * 3 = 9.42 m.

V = (20/9.42) * 2pi = 4.25pi rad/s =
13.34 rad/s.

To solve this problem, we can use the relationship between linear and angular acceleration, as well as the formula for tangential and radial acceleration.

a) To determine the angular velocity (ω) of the hammer, we need to use the formula:

ω = √(α/r),

Where α is the angular acceleration and r is the radius of rotation.
Plugging in the given values: α = 13 rad/s^2 and r = 1.5 m,

ω = √(13 rad/s^2 / 1.5 m) ≈ 2.24 rad/s.

Therefore, the angular velocity of the hammer is approximately 2.24 rad/s.

b) The radial component of linear acceleration (ar) can be calculated using the formula:

ar = rα,

Where α is the angular acceleration and r is the radius of rotation.
Plugging in the given values: α = 13 rad/s^2 and r = 1.5 m,

ar = 1.5 m * 13 rad/s^2 = 19.5 m/s^2.

Therefore, the magnitude of the radial component of the hammer's linear acceleration is 19.5 m/s^2.

c) The tangential component of linear acceleration (at) can be calculated using the formula:

at = rω^2,

Where ω is the angular velocity and r is the radius of rotation.
Plugging in the given values: ω = 2.24 rad/s and r = 1.5 m,

at = 1.5 m * (2.24 rad/s)^2 = 7.56 m/s^2.

Therefore, the magnitude of the tangential component of the hammer's linear acceleration is 7.56 m/s^2.