Answer:
a). linear acceleration of the hoop and block = 3.2895 <u>m</u>
s²
c). magnitude of the angular acceleration of the disk pulley = 41.119 <u>rad</u>
s²
d). tensions in the string between the block and disk pulley = 11.842 N
tensions in the string between the hoop and disk pulley = 15.625 N
check the pictures below for further explanation and for the remaining answers. I hope it helps you. Thank you
Explanation:
Start by writing "F=ma" equations for each of the things that moves. Also, since some of the objects (the pulley and the orange sphere) rotate, you should write "τ = Iα" equations (net torque = moment of inertia × angular acceleration) for those. In the end, you should have enough equations that you can combine them and solve for the desired quantities.
First, the hoop. There's no indication that it rotates, so we don't need a "τ = Iα" equation for it; just do "F=ma". The hoop has gravity ((mhoop)g) pulling down, and the tension in the vertical string (call it "T_v") pulling up.
Fnet = ma
(distance covered) divided by (time to cover the distance) is SPEED.
(straight-line distance and direction between start-point and end-point) is DISPLACEMENT.
(amount and direction of change in speed) divided by (time for the change) is ACCELERATION.
(amount and direction of change in momentum) divided by (time for the change) OR (mass) times (acceleration) is FORCE.
Answer:
angular acceleration =<u>final angular speed + initial angular speed</u>
time
=<u>0.6283rad/s +0</u>
24.6s
=0.6283/24.5
=0.025rad/s^2
Explanation:
first angular speed is equal to angular velocity and we will convert the unit to rad/s so 1rev/sec=6.283rad/s then angular acceleration is the rate at which the angular velocity changes with the time we have the final angular velocity,initial angular velocity and time now substitute the numbers in the formula and the value will be 0.025 rad per seconde square