A 44 m diameter wheel accelerates uniformly about its center from 110 rpm to 320 rpm in 3.9 s.
Determine the radial component of the linear acceleration of a point on the edge of the wheel 1.0 s after it has started accelerating.
Determine the tangential component of the linear acceleration of a point on the edge of the wheel 1.0 after it has started accelerating.
find w at t=1
w=wi+at where a= 2PI/60 (320-110) rad/sec
radial acceleartion= w^2 r
tangential acceleration = a*r
To solve this problem, we can use the formula for linear acceleration:
a = α * r
Where:
a is the linear acceleration
α is the angular acceleration
r is the radius of the wheel
1. To find the radial component of the linear acceleration at a point on the edge of the wheel 1.0 s after it has started accelerating, we need to find the angular acceleration first.
First, convert the initial and final angular velocities from rpm to rad/s:
Initial angular velocity = 110 rpm
Final angular velocity = 320 rpm
Initial angular velocity in rad/s = (110 rpm) * (2π rad/1 min) * (1 min/60 s) = 11.518 rad/s
Final angular velocity in rad/s = (320 rpm) * (2π rad/1 min) * (1 min/60 s) = 33.510 rad/s
Now, find the angular acceleration:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
= (33.510 rad/s - 11.518 rad/s) / 3.9 s
= 5.64 rad/s²
Next, calculate the radial component of the linear acceleration:
Radial linear acceleration = α * r
Given that the diameter of the wheel is 44 m, the radius (r) is half of the diameter:
r = 44 m / 2 = 22 m
Radial linear acceleration = (5.64 rad/s²) * (22 m)
= 124.08 m/s²
Therefore, the radial component of the linear acceleration at a point on the edge of the wheel 1.0 s after it has started accelerating is 124.08 m/s².
2. To find the tangential component of the linear acceleration at a point on the edge of the wheel 1.0 s after it has started accelerating, we can use the formula:
Tangential linear acceleration = α * r
Using the same values for α and r as in the previous calculation:
Tangential linear acceleration = (5.64 rad/s²) * (22 m)
= 124.08 m/s²
Therefore, the tangential component of the linear acceleration at a point on the edge of the wheel 1.0 s after it has started accelerating is 124.08 m/s².
To determine the radial and tangential components of the linear acceleration of a point on the edge of the wheel, we need to use the formula for linear acceleration:
a = ω * r
where a is the linear acceleration, ω is the angular velocity, and r is the radius.
First, let's calculate the initial and final angular velocities.
Given:
Initial angular velocity (ω1) = 110 rpm
Final angular velocity (ω2) = 320 rpm
To convert rpm to radians per second (rad/s), we multiply by (2π/60):
ω1 = (110 rpm) * (2π/60) rad/s ≈ 11.55 rad/s
ω2 = (320 rpm) * (2π/60) rad/s ≈ 33.51 rad/s
Now, let's calculate the change in angular velocity (Δω) using the formula:
Δω = ω2 - ω1
Δω = 33.51 rad/s - 11.55 rad/s ≈ 21.96 rad/s
Next, we need to find the radius (r) of the wheel. Given that the diameter (d) is 44 m, we can calculate the radius (r) using the formula:
r = d/2
r = 44 m/2 = 22 m
Now, let's calculate the linear acceleration using the formula:
a = ω * r
a = Δω/t * r
Since we want to find the linear acceleration 1.0 s after the wheel started accelerating, we use 1.0 s as the time (t) value:
Radial component of linear acceleration (ar) = Δω/t * r
ar = (21.96 rad/s) / (3.9 s) * (22 m) ≈ 124.62 m/s^2
Tangential component of linear acceleration (at) = ω2^2 * r
at = (33.51 rad/s)^2 * 22 m ≈ 24764.47 m/s^2
Therefore, the radial component of the linear acceleration 1.0 s after the wheel started accelerating is approximately 124.62 m/s^2, and the tangential component of the linear acceleration is approximately 24764.47 m/s^2.