A small 0.50 kg block has a horizontal velocity of 3.0 m/s when it slides off a
ID: 1458139 • Letter: A
Question
A small 0.50 kg block has a horizontal velocity of 3.0 m/s when it slides off a 1.2 m high frictionless table as shown in Figure 2. Use the coordinate system with origin at the point Oat the foot of the table. What is the angular momentum of the block about point O at the foot of the table leg (1) just after the block leaves the table and (2) just before the block strikes the floor? What is the torque on the block about point O (1) just after the block leaves the table and (2) just before the block strikes the floor.Explanation / Answer
Displacement along veritcla direction
S =ut+(1/2)at^2
1.2 =0+(1/2)(9.8)t2
t 2= 0.245
t = 0.495 s
v = u+at
vy=0+(9.8*0.495) = 4.851
Displacement along horizontal direction
R = ut
R =(3*0.495) = 1.485 m
(a) m =0.5 kg, R =1.485 m, v =3 m/s
at point (1) L =mvxr = (0.5*3*1.485) =2.23 kg.m2/s
At point (2)
L =mvyr = (0.5*4.851*1.485) =3.602 kg.m2/s
(b) At point (1) the acceleration along x- direction is zero
Torque =R.F = Rmax = 0
At point (2) Torque =Rmay =(1.485*0.5*9.8) = 7.28 N.m