Calculate the maximum and minimum safe speed of vehicle on banked curve road inclined at 37°. If coefficient of friction is 0.5. (Radius of curvature= 250m and g= 10 m/s²).
{Chapter: Circular motion, class: 11}
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Calculate the maximum and minimum safe speed of vehicle on banked curve road inclined at 37°. If coefficient of friction is 0.5. (Radius of curvature= 250m and g= 10 m/s²).
{Chapter: Circular motion, class: 11}
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Verified answer
Answer:
The minimum safe speed is 21.44 m/s.
The maximum safe speed is 70.88 m/s.
Explanation:
Given:
For minimum safe speed,
[tex]\displaystyle{\boxed{\pink{\bf\:v_{min}\:=\:\sqrt{\dfrac{rg\:(\tan\:\theta\:-\:\mu\:)}{1\:+\:\mu\:\tan\:\theta}\:}}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:\sqrt{\dfrac{250\:\times\:10\:(\:\tan\:(\:37^{\circ})\:-\:0.5\:)}{1\:+\:0.5\:\times\:\tan\:(\:37^{\circ}\:)}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:\sqrt{\dfrac{2500\:(\:0.753\:-\:0.5\:)}{1\:+\:0.5\:\times\:0.753}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:\sqrt{\dfrac{2500\:\times\:0.253}{1\:+\:0.3765}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:\sqrt{\dfrac{632.5}{1.3765}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:\sqrt{459.498}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{min}\:=\:21.435}[/tex]
[tex]\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:v_{min}\:\approx\:21.44\:m/s\:}}}}[/tex]
For maximum safe speed,
[tex]\displaystyle{\boxed{\blue{\bf\:v_{max}\:=\:\sqrt{\dfrac{rg\:(\tan\:\theta\:+\:\mu\:)}{1\:-\:\mu\:\tan\:\theta}\:}}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{max}\:=\:\sqrt{\dfrac{250\:\times\:10\:(\:\tan\:(\:37^{\circ})\:+\:0.5\:)}{1\:-\:0.5\:\times\:\tan\:(\:37^{\circ}\:)}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{max}\:=\:\sqrt{\dfrac{2500\:(\:0.753\:+\:0.5\:)}{1\:-\:0.5\:\times\:0.753}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{max}\:=\:\sqrt{\dfrac{2500\:\times\:1.253}{1\:-\:0.3765}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{max}\:=\:\sqrt{\dfrac{3132.5}{0.6235}}}[/tex]
[tex]\displaystyle{\implies\sf\:v_{max}\:=\:\sqrt{5024.057}}[/tex]
[tex]\displaystyle{\therefore\:\underline{\boxed{\purple{\sf\:v_{max}\:=\:70.88\:m/s\:}}}}[/tex]
Explanation:
Answer:
The minimum safe speed is 21.44 m/s.
The maximum safe speed is 70.88 m/s.
Explanation:
Given:
Angle of inclination ( θ ) = 37°
Coefficient of friction
�
μ = 0.5
Radius of curvature ( r ) = 250 m
Acceleration due to gravity ( g ) = 10 m/s²
For minimum safe speed,
�
�
�
�
=
�
�
(
tan
�
−
�
)
1
+
�
tan
�
v
min
=
1+μtanθ
rg(tanθ−μ)
⟹
�
�
�
�
=
250
×
10
(
tan
(
3
7
∘
)
−
0.5
)
1
+
0.5
×
tan
(
3
7
∘
)
⟹v
min
=
1+0.5×tan(37
∘
)
250×10(tan(37
∘
)−0.5)
⟹
�
�
�
�
=
2500
(
0.753
−
0.5
)
1
+
0.5
×
0.753
⟹v
min
=
1+0.5×0.753
2500(0.753−0.5)
⟹
�
�
�
�
=
2500
×
0.253
1
+
0.3765
⟹v
min
=
1+0.3765
2500×0.253
⟹
�
�
�
�
=
632.5
1.3765
⟹v
min
=
1.3765
632.5
⟹
�
�
�
�
=
459.498
⟹v
min
=
459.498
⟹
�
�
�
�
=
21.435
⟹v
min
=21.435
∴
�
�
�
�
≈
21.44
�
/
�
‾
∴
v
min
≈21.44m/s
For maximum safe speed,
�
�
�
�
=
�
�
(
tan
�
+
�
)
1
−
�
tan
�
v
max
=
1−μtanθ
rg(tanθ+μ)
⟹
�
�
�
�
=
250
×
10
(
tan
(
3
7
∘
)
+
0.5
)
1
−
0.5
×
tan
(
3
7
∘
)
⟹v
max
=
1−0.5×tan(37
∘
)
250×10(tan(37
∘
)+0.5)
⟹
�
�
�
�
=
2500
(
0.753
+
0.5
)
1
−
0.5
×
0.753
⟹v
max
=
1−0.5×0.753
2500(0.753+0.5)
⟹
�
�
�
�
=
2500
×
1.253
1
−
0.3765
⟹v
max
=
1−0.3765
2500×1.253
⟹
�
�
�
�
=
3132.5
0.6235
⟹v
max
=
0.6235
3132.5
⟹
�
�
�
�
=
5024.057
⟹v
max
=
5024.057
∴
�
�
�
�
=
70.88
�
/
�
‾
∴
v
max
=70.88m/s