The length of second's hand in a watch is I cm. The change in velocity of its tip in 30 seconds is x pi 30 cm/sec Then I will be
Answers
Answer:
The seconds hand of the watch rotates 2π rad in 60s and its length $$r= 1cm$$
The angular velocity of its tip will be
ω=
60
2π
rad/s
Let the initial position of the tip of the seconds hand be at 12 in the watch. At this position the direction of the velocity of its tip will be towards positive direction of X-axis (considering center of rotation as origin). After 15s the tip rotates
60
2π
∗15=
2
π
rad and the direction of velocity becomes towards negative direction of Y-axis.In both the positions the magnitude of the velocity will be
v=ω∗r=
60
2π
∗1=
30
π
cm/s
Initial velocity vector
v
i
=v
i
^
Final velocity vector
v
f
=−v
j
^
So the change in velocity during 15s will be
⟹
Δv
=
v
f
−
v
i
=−v
j
^
−v
i
^
So magnitude of change in velocity will be
⟹∣
Δv
∣=∣−v
j
^
−v
i
^
∣=
2
v=
30
2
π
cm/s
Answer:
The seconds hand of the watch rotates 2π rad in 60s and its length $$r= 1cm$$
The angular velocity of its tip will be
ω=
60
2π
rad/s
Let the initial position of the tip of the seconds hand be at 12 in the watch. At this position the direction of the velocity of its tip will be towards positive direction of X-axis (considering center of rotation as origin). After 15s the tip rotates
60
2π
∗15=
2
π
rad and the direction of velocity becomes towards negative direction of Y-axis.In both the positions the magnitude of the velocity will be
v=ω∗r=
60
2π
∗1=
30
π
cm/s
Initial velocity vector
v
i
=v
i
^
Final velocity vector
v
f
=−v
j
^
So the change in velocity during 15s will be
⟹
Δv
=
v
f
−
v
i
=−v
j
^
−v
i
^
So magnitude of change in velocity will be
⟹∣
Δv
∣=∣−v
j
^
−v
i
^
∣=
2
v=
30
2
π
cm/s