BRYCE Program Project

Physics Problems 4.

a) 2:95 rad/s �1:9 m=5:605 m/s ac =

v2

r =

(5:605)2

1:9 = 16:535 m/s^2

b) Ffriction = Fc(Force of friction must equal the centripedal force) Ffriction = mac = 50 �16:535 = 826:75 N

c) Ffriction = 826:75 �mg = 826:75 � = 826:75

mg = 826:75

50�9:8 = 1:6872

Since the coe¢ cient of friction is greater than 1, this situation is not possible

5. a) U = �MearthmG

rearth+h = �5:97�10

24�96�6:67�10�11 6371000+1:97�106 = �4:583�10

9 Joules

b) Fg =

MearthmG r2

= 5:97�10 24�96�6:67�10�11

(6371000+1:97�106)2 = 549:46 N

7. a) Fg = mac

MmG (rearth+h)

2 = m v2

rearth+h MG

rearth+h = v2

v = d t =

2�(rearth+h) t

MG rearth+h

= � 2�(rearth+h)

t

�2 MG

rearth+h =

4�2(rearth+h) 2

t2

(rearth +h) 3 = MGt

2

4�2

h = 3 q

MGt2

4�2 �rearth =

3

q 5:97�1024�6:67�10�11(103�60)2

4�2 �6371000 = 9:0522�105

m

b) Fg = mg

MmG (rearth+h)

2 = mg

1

g = MG (rearth+h)

2 = 5:97�1024�6:67�10�11 (6371000+9:0928�105)2 = 7:5128 m/s^2

8. a) Fg = mac MmG

(2rearth) 2 = m

v2

2rearth

v2 = MG (2rearth)

v = q

MG (2rearth)

= q

5:97�1024�6:67�10�11 2�6371000 = 5590:3 m/s

b) t = d

v = 2�2rearth

v = 4��6371000

5590:3 = 14321 s

14321 3600

= 3:9781 s

c) Fg =

MmG (2rearth)

2 = 5:97�1024�788�6:67�10�11

(2�6371000)2 = 1932:6 N

2