A particle is following a circular path and is observed to have an angular displacement of 10.3°.

 Question 3

A particle is following a circular path and is observed to have an angular displacement of 10.3°.

(a) Express this angle in radians (rad). Show your working and give your answer to three significant figures. [2]

 

 

(b) (i) Determine tan10.3° to three significant figures.

 

(ii) Hence calculate the percentage error that is made when the angle 10.3°, as measured in radians, is assumed to be equal to tan10.3°.

[3]

 

 

Reference: Past Exam Paper – November 2004 Paper 4 Q1

 

 

Solution:

(a)

 θ (rad) = 2π × (10.3/360)                    

θ (rad) = 0.180 rad                             

 

 

(b) (i) tan θ = 0.182                            

 

(ii)

{Difference between θ and tanθ = 0.182 – 0.180 = 0.002}

 

percentage error = (0.002/0.180) x 100                                

percentage error = 1.1 (%)                                                    

A steel wire of cross-sectional area 15 mm2 has an ultimate tensile stress of 4.5 × 108 N m-2.

Question 21

A steel wire of cross-sectional area 15 mm² has an ultimate tensile stress of 4.5 × 10⁸ N m^-2.

(a) Calculate the maximum tension that can be applied to the wire. [2]

 

 

(b) The steel of the wire has density 7800 kg m^-3. The wire is hung vertically.

 

Calculate the maximum length of the steel wire that could be hung vertically before the wire breaks under its own weight. [3]

 

 

 

Reference: Past Exam Paper – November 2015 Paper 23 Q7

 

 

 

Solution:

(a)

{Stress = Force / Area

Note that the tension is a force.}

 

stress σ = F / A                       

 

{Convert area into m² by multiplying by 10^-6.

Tension = Stress × Area}

max. tension = UTS × A = 4.5×10⁸ × 15×10^-6 = 6800 (6750) N

 

 

(b)

{We first need to relate the length with data available.}

Density ρ = m / V                               

 

weight = mg = ρVg = ρALg                 {since Volume V = Area × Length = AL}

 

{When the wire break under its own weight, the weight should be equal to the maximum tension.

Weight = ρALg

Tension = ρALg}

6750 = 7.8×10³ × 15×10^-6 × L × 9.81             

L = 5.9 (5.88) × 10³ m