Respuesta :
Answer:
A. α = 94.4 rad/s
B. a = 28.32 m/s
C. N = 34N
D. α = 94.4 rad/s
a = 28.32 m/s
N = 44.4 N
Explanation:
part A:
using:
∑T = Iα
where T is the torque, I is the moment of inertia and α is the angular momentum.
firt we will find the moment of inertia I as:
I = [tex]\frac{1}{2}MR^2[/tex]
Where M is the mass and R is the radius of the wheel, then:
I = [tex]\frac{1}{2}(8 kg)(0.3 m)^2[/tex]
I = 0.36 kg*m^2
Replacing on the initial equation and solving for α, we get::
∑T = Iα
Fr = Iα
34 N = 0.36α
α = 94.4 rad/s
part B
we need to use this equation :
a = αr
where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:
a = (94.4)(0.3 m)
a = 28.32 m/s
part C
Using the laws of newton, we know that:
N = T
where N is the force that the axle exerts on the wheel part and T is the tension of the cord
so:
N = 34N
part D
The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:
-Mg + N +T = 0
Then, solving for N, we get:
N = -T+Mg
N = -34 + (8 kg)(9.8)
N = 44.4 N
