A 4.00kg counterweight is attached to a light cord, which is would around a spool. The spool is a uniform solid cylinder of radius 8.00cm and mass 2.00kg. (a) What is the net torque on the system about the point O (the origin)? (b) When the counterweight has a speed v, the pulley has an angular speed ω=v/R. Determine the total angular momentum of the system about O. (c) Using the fact that τ=dL/dt and your result from (b), calculate the acceleration of the counterweight. slader

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

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

false

explanation:

boats, things on boats luggage are way more heavy then water.