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A 2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is 胃 = 30颅o.
a. What is the ball鈥檚 speed when 胃 = 30o颅?
b. What are the magnitude and direction of the ball鈥檚 acceleration when 胃 = 30o颅?

Short Answer

Expert verified

a) Balls speed is 3.85 m /sec

b) Ball's acceleration is 18.496 m/sec2

Step by step solution

01

Part(a) Step 1 : Given Information

Mass of ball is 2 kg
Length of string = 80 cm =0.8 m
Tension in the string = 20 N
When its angle from the highest point on the circle is 胃 = 30o颅.

02

Part(a) Step 2: Explanation

The centripetal force acting on a mass m revolving with speed v around a circle of radius r is given by

mv2/r 鈥︹︹︹︹︹︹︹︹︹︹︹.(1)

The centripetal acceleration is given by

v2/r 鈥︹︹︹︹︹︹︹︹︹︹︹.(2)

The tension of the spring at a given angle is and solve for v

T=mv2r-mgcos(30)mv2r=T+mgcos(30)v2=r(T+mgcos(30))mv=r(T+mgcos(30))m

Substitute the given value we get

v=(0.8m)(20N)+(2kg)(9.8m/s2)cos(30)2kg=3.85m/s

03

Part(b) Step 1: Given information

Mass of ball is 2 kg
Length of string = 80 cm =0.8 m
Tension in the string = 20 N
When its angle from the highest point on the circle is 胃 = 30o颅.

04

Part(b) Step 2: Explanation

The ball's radial acceleration is calculated as

a=v2r

Substitute the values given

a=(3.85m/s)2(0.8m)=18.496m/s2

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