A solid sphere of radius r rolls without slipping from rest from a height h of an inclined
track at the bottom of which there is a loop of radius R much larger than the radius of
sphere, as shown in figure. The minimum value of h for the sphere to complete the loop is
Answers
Answer:
user_photo
Type your question...
11th
Physics
Systems of Particles and Rotational Motion
Rolling Motion
A small solid sphere of rad...
PHYSICS
A small solid sphere of radius r rolls down an incline without slipping which ends into a vertical loop of radius R. Find the height above the base so that it just loops the loop
November 22, 2019avatar
Gurmeet Ñegī
SHARE
ANSWER
Let v
1
and w be the velocity and angular velocity of the sphere at B respectively.
As the is no friction. so conservation of energy is valid.
P.E
A
+K.E
A
=P.E
B
+K.E
B
mgh+0=mg(2R−r)+
2
1
mv
1
2
+
2
1
Iw
2
mgh=mg(2R−r)+
2
1
mv
1
2
+
2
1
×
5
2
mr
2
w
2
mgh−mg(2R−r)=
2
1
mv
1
2
+
5
1
mv
1
2
[v
1
=rw]
⟹mg(h+r−2R)=
10
7
mv
1
2
Circular motion equation:
R−r
mv
1
2
=mg
⟹mg(h+r−2R)=
10
7
mg(R−r)
⟹h=
10
27
R−
10
17
r
Explanation:
To just complete the track potential energy at height h
(Point P)=(Kinetic+ potential) energy at H
mgh=2mgR+
2
1
mv
0
2
+
2
1
Iω
0
2
where v
0
is velocity required to complete the loop
R
mv
0
2
=mg⇒v
0
2
=gR
mgh=2mgR+
2
1
mv
0
2
+
2
1
×
5
2
mr
2
×
R
v
0
2
gh=2gR+
2
1
gR+
5
1
R
2
r
2
.gR (ω
0
=
R
v
0
pure rolling)
h=
2
5R
+
5
R
[
R
2
r
2
]