Questions from chapter 8.rotational of a rigid body

the moment of inertia of wheel= 1/2mr²

m= mass of wheel= 2.8 kg

r= radius

i= 1/2 (2.8)(0.35)²=0.1715 kgm²

(a) torque = F r

F= force=5 N

r= radius=0.35 m

τ= 5 (0.35)

τ= 1.75 Nm

(b) for angular acceleration, we use

τ= iα

τ= torque= 1.75 Nm

i= moment of inertia=0.1715 kgm²

so 1.75 =0.1715 α

α=10.2 rad/s²

(c) for angular displacement θ, we use the kinematic equation,

θ= ωo t + 1/2α t²

ωo= initial angular velocity=0

t= time=4 s

α= angular acceleration=10.2 rad/s²

so θ = 0 (4) + 1/2 (10.2)(4)²

θ=81.6 rad

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be