Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s
Answer:
a) τnet = 3.1392 N-m
b) L = (0.48 Kg-m)*v
c) a = 6.54 m/s²
Explanation:
Given
m = 4 Kg
R = 8 cm = 0.08 m
M = 2 Kg
a) τnet = ?
b) L = ?
c) a = ?
Solution
a) We use the formula
τnet = R*m*g*Sin 90°
τnet = 0.08 m*4 Kg*9.81 m/s²*(1)
τnet = 3.1392 N-m
b) We apply the equation
L= R*m*v + R*M*v = R*(m + M)*v
then
L = (0.08 m)*(4 Kg + 2 Kg)*v = (0.48 Kg-m)*v
c) We use the relation
τ = dL/dt = d((0.48 Kg-m)*v)/dt = (0.48 Kg-m)*dv/dt
τ = (0.48 Kg-m)*a
then
τ/(0.48 Kg-m) = a
⇒ a = 3.1392 N-m/((0.48 Kg-m)
a = 6.54 m/s²
