# Physics 9702 Doubts | Help Page 177

__Question 877: [Circular motion]__
A large bowl is made from part of a
hollow sphere.

A small spherical ball is placed
inside the bowl and is given a horizontal speed. The ball follows a horizontal
circular path of constant radius, as shown in Fig.1.

The forces acting on the ball are
its weight W and the normal reaction force R of the bowl on the ball, as shown
in Fig.2.

The normal reaction force R is at an
angle θ to the horizontal.

**(a)**

(i) By resolving reaction force R
into two perpendicular components, show that the resultant force F acting on
the ball is given by the expression

W = F tan θ.

(ii) State significance of the force
F for the motion of the ball in the bowl.

**(b)**The ball moves in circular path of radius 14 cm. For this radius, the angle θ is 28°. Calculate the speed of the ball.

**Reference:**

*Past Exam Paper – November 2014 Paper 41 & 42 Q2*

__Solution 877:__**(a)**

(i)

{The weight acts
vertically down, so it does not have any horizontal component. The force
component of the force F is then only due to the horizontal component of the
reaction force R.}

F = R cosθ

{The path is circular with
a constant radius. This means that the ball does not move up or down in the
surface of the sphere. Thus, the resultant vertical component of the force on
the ball is zero – that is, the vertical upward component of the reaction force
R is equal to the downward weight.}

W = R sinθ

{Dividing the equations
eliminate R and gives the required expression.}

dividing, W = F tanθ

(ii) The force F

__provides__the centripetal force.**(b)**

{As shown above, W = F
tanθ. The weight W is equal to mg. The resultant force F provides the
centripetal force, given by mv

^{2}/ r. Replacing these in the equation (W = F tanθ) allows us to solve for speed v.}
EITHER F = mv

^{2}/ r and W = mg
{W = F tanθ giving mg = (mv

^{2}/ r) tan θ}
OR v

^{2}= rg / tan θ
v

^{2}= (14×10^{–2}× 9.8) / tan 28° = 2.58
Speed v = 1.6 m s

^{–1}

__Question 878:__
A thermometer can be read to
accuracy of ± 0.5 °C. This thermometer is used to measure a temperature rise
from 40 °C to 100 °C.

What is percentage uncertainty in
the measurement of the temperature rise?

A 0.5 % B 0.8 % C
1.3 % D 1.7 %

**Reference:**

*Past Exam Paper – November 2003 Paper 1 Q4 & June 2014 Paper 12 Q5*

__Solution 878:__**Answer: D.**

Temperature rise = 100 – 40 = 60°C

The ±
0.5°C uncertainty happens at the measurement of both the low and
the high temperatures.

Percentage uncertainty in
measurement = [ΔT / T] × 100%
= [2(0.5) / 60] × 100% = 1.7%

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ReplyDeleteFor question 4, see solution 880 at

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ReplyDeleteFor 4/M/J/02 Q.7(b)(c), see solution 882 at

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