Question : An electric bulb suspended from a roof of a railway train by a flexible wire shifts through an angle of 19°48', when the train goes horizontally round a curved path of 200 m radius. Find the speed of the train.
Doubt by Muskan
Solution :
θ = 19°48'
= 19° + (48/60)°
= 19° + (4/5)°
= 19° + 0.8°
= 19.8°
= 19° + (48/60)°
= 19° + (4/5)°
= 19° + 0.8°
= 19.8°
r = 200 m
The various forces acting on the bulb are shown in the figure. For simplicity the bulb is represented as a black dot.
Here
T = Tension in the string
θ = Angle made by the string with the vertical.
F = Centripetal Force
mg = Weight of bulb acting vertically downward.
θ = Angle made by the string with the vertical.
F = Centripetal Force
mg = Weight of bulb acting vertically downward.
Resolving the Tension into two rectangular components Tcosθ and Tsinθ as shown in figure.
At equilibrium
Tsinθ = mv2/r ------ (1)
Tcosθ = mg --------(2)
Dividing equation (1) by (2)
Tsinθ/Tcosθ = mv2/mgr
tanθ = v2/rg
v2=rgtanθ
v = √(rgtanθ)
Tcosθ = mg --------(2)
Dividing equation (1) by (2)
Tsinθ/Tcosθ = mv2/mgr
tanθ = v2/rg
v2=rgtanθ
v = √(rgtanθ)
v=√[200×9.8×tan(19.8°)]
v=√[200×9.8×0.36] [∵tan 19.8° = 0.36]
v=√705.6
v= 26.5631 m/s
v= 26.56 m/s
v= 26.5631 m/s
v= 26.56 m/s